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Example 1. Find $\lim _{x \rightarrow 2} \frac{x^{2}-3 x+2}{\left|x^{2}-6 x+8\right|}=\left(\frac{0}{0}\right)$. | Solution.
$$
\begin{aligned}
& \lim _{x \rightarrow 2-0} \frac{(x-1)(x-2)}{|x-2||x-4|}=\lim _{x \rightarrow 2} \frac{(x-1)(x-2)}{-(x-2)|x-4|}=-\frac{1}{2} \\
& \lim _{x \rightarrow 2+0} \frac{(x-1)(x-2)}{|x-2||x-4|}=\lim _{x \rightarrow 2} \frac{(x-1)(x-2)}{(x-2)|x-4|}=\frac{1}{2}
\end{aligned}
$$
The left-hand and r... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,217 |
Example 2. Find $\lim _{x \rightarrow 1} 3^{(2+x) /(x-1)}$. | Solution.
$\lim _{x \rightarrow+1-0} 3^{(2+x) /(x-1)}=\left\{\begin{array}{l}\text { the exponent in absolute value increases, } \\ \text { but its sign is negative; the value under } \\ \text { the limit sign decreases, remaining } \\ \text { positive }\end{array}\right\}=0 ;$
$\lim _{x \rightarrow 1+0} 3^{(2+x) /(x-... | doesnotexist | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,218 |
Example 3. Find $\lim _{x \rightarrow 1}\left(\frac{2 x^{2}+3 x-5}{x^{2}-1}\right)^{2}=\left(\frac{0}{0}\right)$. | Solution.
$$
\begin{aligned}
& \lim _{x \rightarrow 1-0}\left(\frac{(x-1)(2 x+5)}{(x-1)(x+1)}\right)^{2}=\left(\frac{7}{2}\right)^{2}=\frac{49}{4} \\
& \lim _{x \rightarrow 1+0}\left(\frac{(x-1)(2 x+5)}{(x-1)(x+1)}\right)^{2}=\left(\frac{7}{2}\right)^{2}=\frac{49}{4}
\end{aligned}
$$
Therefore, $\lim _{x \rightarrow ... | \frac{49}{4} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,219 |
Example 4. Let $f(x)=\left\{\begin{array}{c}-x+1 \text { for } x \leqslant 2, \\ 2 x+1 \text { for } x>2 .\end{array}\right.$ Find $f(2+0)$ and $f(2-0)$ : | Solution.
$$
\begin{gathered}
f(2+0)=\lim _{x \rightarrow 2+0} f(x)=\lim _{x \rightarrow 2}(2 x+1)=5 \\
f(2-0)=\lim _{x \rightarrow 2-0} f(x)=\lim _{x \rightarrow 2}(-x+1)=-1
\end{gathered}
$$
## Exercises
Find the right and left limits of the functions.
1. $f(x)=\frac{1}{5+3^{1 / x}}$ as $x \rightarrow 0$.
2. $f(x... | f(2+0)=5,f(2-0)=-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,220 |
Example 3. Investigate the continuity of the function
$$
f(x)=\left\{\begin{array}{l}
\frac{x^{2}+7 x+10}{x^{2}-4} \quad \text { for } \quad x<2, x \neq-2 \\
\frac{11}{4} \text { for } \quad x=-2 \\
4 x-3 \quad \text { for } \quad x \geqslant 2
\end{array}\right.
$$ | Solution. First, let's investigate the continuity at the point $x=-2$. We have:
$$
\begin{gathered}
\lim _{x \rightarrow-2} f(x)=\lim _{x \rightarrow-2} \frac{x^{2}+7 x+10}{x^{2}-4}=\left(\frac{0}{0}\right)=\lim _{x \rightarrow-2} \frac{(x+2)(x+5)}{(x+2)(x-2)}=-\frac{3}{4} \\
f(-2)=\frac{11}{4} \neq \lim _{x \rightarr... | \ | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,223 |
Example 4. Investigate the continuity and plot the graph of the function
\[
f(x)=\left\{\begin{array}{l}
x+1.5 \quad \text { for } \quad x<-2 \\
\frac{1}{x} \quad \text { for } \quad-2 \leqslant x<0 \\
2 x \quad \text { for } \quad x \geqslant 0
\end{array}\right.
\] | Solution. The function $f(x)$ is defined on the entire set of real numbers and is given by three expressions on different intervals of argument variation.
Let's investigate the continuity of the function at the points $x=-2$ and $x=0$:
$$
f(-2-0)=\lim _{x \rightarrow-2-0}(x+1.5)=-0.5 ; \quad f(-2+0)=\lim _{x \rightar... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,224 |
Example 5. It is required to determine whether the given function is continuous or discontinuous for each of the given argument values; in the case of a discontinuity, find the limit at the point of discontinuity from the right and from the left and make a schematic graph of the function
$$
f(x)=4^{1 /(3-x)}, \quad x_... | Solution. The function $f(x)$ is elementary, which means it is continuous at any point in its domain. At the point $x_{1}=1$, this function is defined, so it is continuous.
At the point $x_{2}=3$, the function is undefined, so it is discontinuous. Let's determine the nature of the discontinuity. We find
$$
\lim _{x \... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,225 |
## Example 1. Given numbers:
a) $z_{1}=2-i$, b) $z_{2}=-\sqrt{7}\left[\cos \left(\operatorname{arctg} \frac{\sqrt{3}}{2}\right)+i \sin \left(\operatorname{arctg} \frac{\sqrt{3}}{2}\right)\right]$, c) $z_{3}=-2 \sqrt{3}(1-i)$, d) $z_{4}=\frac{1}{14 \sqrt{30}} e^{-2 \operatorname{arctg}(\sqrt{3} / 2) i}$.
It is require... | Solution. a) $\left|z_{1}\right|=\sqrt{4+1}=\sqrt{5}, \arg z_{1}=\operatorname{arctg}\left(-\frac{1}{2}\right)=-\operatorname{arctg} \frac{1}{2}$, $z_{1}=\sqrt{5}\left[\cos \left(-\operatorname{arctg} \frac{1}{2}\right)+i \sin \left(-\operatorname{arctg} \frac{1}{2}\right)\right]=\sqrt{5} e^{-i \operatorname{arctg}(1 /... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,226 |
Example 1. Decompose into partial fractions: $\frac{1}{(x+1)\left(x^{2}+1\right)}$. | ## Solution. We have
$$
\frac{1}{(x+1)\left(x^{2}+1\right)}=\frac{A}{x+1}+\frac{B x+C}{x^{2}+1}=\frac{A\left(x^{2}+1\right)+(B x+C)(x+1)}{(x+1)\left(x^{2}+1\right)}
$$
Apply Theorem 3. We arrive at the equality $1=A x^{2}+A+B x^{2}+B x+$ $+C x+C$ (the denominators of the equal fractions are the same). Now apply Theor... | \frac{1}{2(x+1)}+\frac{-x+1}{x^{2}+1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,228 |
Example 2. Decompose into partial fractions: $\frac{1}{(x+1)(x+2)^{2}(x+3)^{3}}$. Solution. According to Theorem 2, we have the decomposition
$$
\begin{aligned}
& \frac{1}{(x+1)(x+2)^{2}(x+3)^{3}}= \\
& \quad=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{(x+2)^{2}}+\frac{D}{(x+3)}+\frac{E}{(x+3)^{2}}+\frac{F}{(x+3)^{3}}
\end{a... | Solution.
1) The given fraction is improper, so we first separate the integer part.
$$
\begin{aligned}
& 2 x^{4}-13 x^{3}+32 x^{2}-24 x+1 \left\lvert\, \frac{x^{3}-5 x^{2}+6 x}{2 x-3}\right. \\
& \frac{2 x^{4}-10 x^{3}+12 x^{2}}{-3 x^{3}+20 x^{2}-24 x+1} \\
& \quad \frac{-3 x^{3}+15 x^{2}-18 x}{5 x^{2}-6 x+1}
\end{al... | \frac{1}{(x+1)(x+2)^{2}(x+3)^{3}}=\frac{1}{8(x+1)}+\frac{2}{x+2}-\frac{1}{(x+2)^{2}}-\frac{17}{8(x+3)}-\frac{5}{4(x+3)^{2}}-\frac{1}{2(x+3)^{3}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,229 |
Example 1. Find the equations of the tangent and normal to the parabola $y=x^{2}+4 x+2$ at the point $A(1,7)$. | Solution. The point $A(1,7)$ lies on the parabola. We find $y^{\prime}=2 x+4$ and $y^{\prime}(1)=6$. Therefore (see point $1^{\circ}$), the equation of the tangent line is $y=7+6(x-1)$, i.e., $y=6 x+1$, and the equation of the normal (see point $2^{\circ}$) is $y=7-\frac{1}{6}(x-1)$, i.e., $x+6 y-43=0$. | 6x+1,\;x+6y-43=0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,230 |
Example 2. Form the equation of the line passing through the point $B(-1,2)$ and tangent to the parabola $y=x^{2}+4 x+9$. | Solution. Point $B(-1,2)$ does not lie on the parabola, so the method described above is not applicable. Let $M_{0}\left(x_{0}, y_{0}\right)$ be the point of tangency of the sought tangent line with the parabola. Then $k_{0}=2 x_{0}+4-$ is the slope of this line. The equality $y_{0}=2+\left(2 x_{0}+4\right)\left(x_{0}+... | -2x,6x+8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,231 |
Example 1. $y=x^{10}-\frac{1}{5} x^{5}+7$ | Solution. Applying rules 1) and 2), as well as formulas 3) and 1): $y^{\prime}=\left(x^{10}\right)^{\prime}-\left(\frac{1}{5} x^{5}\right)^{\prime}+(7)^{\prime}=10 x^{9}-\frac{1}{5} \cdot 5 x^{4}+0=10 x^{9}-x^{4}$. | 10x^{9}-x^{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,232 |
Example 3. $y=2^{x} \cdot \cos x$. | Solution. We have the derivative of the product (rule 3)) of functions contained in formulas 4) and 9): $y^{\prime}=\left(2^{x}\right)^{\prime} \cos x+2^{x}(\cos x)^{\prime}=$ $=2^{x} \ln 2 \cos x-2^{x} \sin x$ | 2^{x}\ln2\cosx-2^{x}\sinx | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,233 |
Example 4. $y=\frac{\tan x-e^{x}}{\sqrt{x}}$. | Solution. Apply rule 4), formulas 10), 5), 16). The derivative of the quotient:
$$
y^{\prime}=\frac{\left(\frac{1}{\cos ^{2} x}-e^{x}\right) \sqrt{x}-\left(\operatorname{tg} x-e^{x}\right) \frac{1}{2 \sqrt{x}}}{x}
$$ | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,234 |
Example 6. $y=\operatorname{arctg} \sqrt{x}$. | Solution. We apply rule 5), formulas 14) and 16). Let $u=\sqrt{x}$. Then $y=\operatorname{arctg} u, y_{u}^{\prime}=\frac{1}{1+u^{2}}, u^{\prime}=\frac{1}{2 \sqrt{x}}$. Finally: $y_{u}^{\prime}=\frac{1}{2 \sqrt{x}(1+x)}$. | \frac{1}{2\sqrt{x}(1+x)} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,236 |
Example 7. $y=\ln \left(x+\sqrt{x^{2}+1}\right)$. | Solution. Let $y=\ln u, u=x+\sqrt{x^{2}+1}$, and apply rule 5). We get $y^{\prime}=\frac{1}{x+\sqrt{x^{2}+1}}\left(1+\frac{2 x}{2 \sqrt{x^{2}+1}}\right)=\frac{1}{\sqrt{x^{2}+1}}$. | \frac{1}{\sqrt{x^{2}+1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,237 |
Example 8. $y=\cos \ln x$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
Example 8. $y=\cos \ln x$. | Solution. Let $y=\cos u, u=\ln x$. We find $y^{\prime}=-\sin \ln x \times$ $\times \frac{1}{x}=-\frac{\sin \ln x}{x}$ | -\frac{\sin\lnx}{x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,238 |
Example 9. $y=\frac{\cos ^{2} 2 x}{\tan x^{2}}$. | Solution. Differentiate the quotient of two complex functions:
$$
\begin{aligned}
y^{\prime}= & \frac{\left(\cos ^{2} 2 x\right)^{\prime} \cdot \operatorname{tg} x^{2}-\left(\operatorname{tg} x^{2}\right)^{\prime} \cdot \cos ^{2} 2 x}{\left(\operatorname{tg} x^{2}\right)^{2}}= \\
& =\frac{2 \cos 2 x \cdot(-\sin 2 x) \... | \frac{-\sin4x\cdot\sin2x^{2}-2x\cos^{2}2x}{\sin^{2}x^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,239 |
Example 10. $y=\sin ^{3} x^{2} \cdot \arccos \sqrt{x}$. | Solution. We are dealing with the derivative of a product of composite functions:
$$
\begin{aligned}
& y^{\prime}=\left(\sin ^{3} x^{2}\right)^{\prime} \cdot \arccos \sqrt{x}+\sin ^{3} x^{2}(\arccos \sqrt{x})^{\prime}= \\
& =3 \sin ^{2} x^{2} \cdot \cos x^{2} \cdot 2 x \cdot \arccos \sqrt{x}-\frac{\sin ^{3} x^{2}}{\sq... | 6x\sin^{2}x^{2}\cdot\cosx^{2}\arccos\sqrt{x}-\frac{\sin^{3}x^{2}}{2\sqrt{x(1-x)}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,240 |
Example 11. $y=\frac{(x+2)^{2}}{(x-1)^{3}(x+5)^{4}}$. | Solution. First, we find $\ln y$, then differentiate both sides of the obtained equation, where the left side $\ln y(x)$ is differentiated as a composite function: $\ln y=2 \ln (x+2)-3 \ln (x-1)-4 \ln (x+5)$ and $\frac{1}{y} y^{\prime}=\frac{2}{x+2}-\frac{3}{x-1}-\frac{4}{x+5} ;(\ln y)^{\prime}=\frac{1}{y} y^{\prime}-$... | y^{\}=\frac{(x+2)^{2}}{(x-1)^{3}(x+5)^{4}}\cdot(\frac{2}{x+2}-\frac{3}{x-1}-\frac{4}{x+5}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,241 |
Example 12. $y=(1+x)^{x}$. | Solution. Take the logarithm of both sides of the equality $y=(1+x)^{x}$: $\ln y=x \ln (x+1)$. Differentiate: $\frac{1}{y} y'=\ln (x+1)+\frac{x}{x+1}$. From this, $y'=(1+x)^{x}\left[\ln (x+1)+\frac{x}{x+1}\right]$.
## Exercises
## Find the derivatives of the functions.
1. $y=\frac{x^{2}-1}{x^{2}+1} . \quad$ 2. $y=\s... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,242 |
Example 1. Calculate approximately the value of the function $f(x)=$ $=\sqrt{x^{2}+x+3}$ at $x=1.97$ | Solution. Let $x_{0}=2$. Then $y_{0}=f(2)=\sqrt{2^{2}+2+3}=3$. Further,
$$
y^{\prime}=\left.\frac{2 x+1}{2 \sqrt{x^{2}+x+3}}\right|_{x=2}=\frac{5}{6}
$$
Since $\Delta x=x-x_{0}$, then $\Delta x=-0.03$ and, according to the formula in point $3^{\circ}$, we have $\sqrt{1.97^{2}+1.97+3} \approx 3+\frac{5}{6}(-0.03)=2.97... | 2.975 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,243 |
Example 2. Approximately calculate the increase in the volume of a cylinder with height $H=40 \mathrm{~cm}$ and base radius $R=30 \mathrm{~cm}$ when the radius is increased by $0.5 \mathrm{~cm}$. | Solution. The volume of a cylinder $V$ with a constant height $H$ and a variable radius $x$ is a function of $x: V=\pi H x^{2}$. For $x=R=30$, we have $V=36000 \pi$. The increment of the cylinder's volume is replaced by its differential:
$$
\Delta V \approx d V=2 \pi H x \cdot \Delta x=2 \pi \cdot 40 \cdot 30 \cdot 0.... | \DeltaV=1200\pi\mathrm{}^{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,244 |
Example 3. Find the differential of the function $y=e^{x}\left(x^{2}+3\right)$. Calculate the value of the differential at the point $x=0$. | Solution. We have:
$$
\begin{gathered}
d y=y^{\prime} d x=\left(e^{x}\left(x^{2}+3\right)+2 x e^{x}\right) d x=\left(x^{2}+2 x+3\right) e^{x} d x \\
d y(0)=\left.\left(x^{2}+2 x+3\right) e^{x}\right|_{x=0} d x=3 d x
\end{gathered}
$$
## Exercises
Find the differentials of the functions.
1. $y=\ln \left(\frac{1+x}{1... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,245 |
Example 1 (to $4^{\circ}$). Find $y^{\prime}$ if $10 x^{3}+4 x^{2} y+y^{2}=0$. Also find $y^{\prime}$ at $x=-2, y=4$. | Solution. Assume that $y=y(x)$. Differentiate the left-hand side of the given equation as a composite function, set the obtained expression to zero, and find $y^{\prime}$:
$$
30 x^{2}+8 x y+4 x^{2} y^{\prime}+2 y y^{\prime}=0
$$
From this, $y^{\prime}=\frac{-15 x^{2}-4 x y}{2 x^{2}+y}$. Substituting $x=-2, y=4$, we g... | -\frac{7}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,246 |
Example 2 (to $4^{\circ}$). Find $y^{\prime}$ if $e^{y}+x y=e$. Also find $y^{\prime}(0)$. | Solution. Differentiate both sides of the given equality, keeping in mind that $y=y(x): e^{y} y^{\prime}+y+x y^{\prime}=0$, from which we find $y^{\prime}=-\frac{y}{e^{y}+x}$. In particular, if $x=0$, then $y=1$ and $y^{\prime}(0)=-\frac{1}{e}$. | y^{\}(0)=-\frac{1}{e} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,247 |
Example 3 (to item $4^{\circ}$). For the function $y(x)$ defined implicitly by the equation $y e^{x}+e^{y}=e+1$, find $y^{\prime \prime}$. | Solution. After two consecutive differentiations of the given equation with respect to $y=y(x)$, we obtain
$$
y^{\prime} e^{x}+e^{x} y+e^{y} y^{\prime}=0
$$
$$
y^{\prime \prime} e^{x}+e^{x} y^{\prime}+e^{x} y+e^{x} y^{\prime}+e^{y}\left(y^{\prime}\right)^{2}+e^{y} y^{\prime \prime}=0
$$
From the second equality, we ... | y^{\\}=\frac{-2e^{2x}y(e^{x}+e^{y})+e^{x}y(e^{x}+e^{y})^{2}+y^{2}e^{y}e^{2x}}{(e^{x}+e^{y})^{3}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,248 |
Example 5 (to $5^{\circ}$ ). Find $y^{\prime \prime}$, if $x=\ln t, y=\sin 2 t$. | Solution. To determine $y^{\prime \prime}$, we will use the second formula. First, we find: $x_{t}^{\prime}=\frac{1}{t}, x_{t^{2}}^{\prime \prime}=-\frac{1}{t^{2}}, y_{t}^{\prime}=2 \cos 2 t, y_{t^{2}}^{\prime \prime}=-4 \sin 2 t$.
Therefore,
$$
y_{x^{2}}^{\prime \prime}=\frac{-4 \sin 2 t \cdot \frac{1}{t}+\frac{1}{t... | -4^{2}\cdot\sin2+2\cos2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,249 |
Example 1. Check if Rolle's theorem is valid for the function $f(x)=x^{2}+2 x+7$ on the interval $[-6 ; 4]$, and if so, find the corresponding value of c. | Solution. $f(x)=x^{2}+2 x+7$ is continuous and differentiable on any interval, in particular, on $[-6 ; 4]$, and $f(6)=f(4)=31$. Therefore, $f^{\prime}(x)=2 x+2=0$ for some $x \in(-6 ; 4)$. We have $x=c=-1$. | -1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,250 |
Example 2. Find the point $c$ mentioned in Lagrange's theorem for the function $y=x^{2}+6 x+1$ on the interval $[-1 ; 3]$. | Solution. We have $y(-1)=-4, y(3)=28, A(-1,-4), B(3,28)$, $k_{A B}=\frac{28+4}{3+1}=8$. We need to solve the equation $y^{\prime}=2 x+6=8$. We find $x=1$, i.e., $C$ has coordinates $(1,8)$. The tangent at this point is parallel to the chord $A B$.
## Exercises
Check whether Rolle's theorem is applicable to the given ... | \frac{1}{} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,251 |
Example 1. Derive the equations of the tangent and normal lines to the curve $y=x \ln x$, drawn at the point with abscissa $x=e$. | Solution. We have: $x_{0}=e, y_{0}=y(e)=e, y^{\prime}=\ln x+1, k=y^{\prime}(e)=2$,
$(t): y=e+2(x-e),(n): y=e-\frac{1}{2}(x-e)(\text{cm} . \S 2)$.
Answer. $(t): y=2 x-e,(n): y=-\frac{1}{2} x+\frac{3}{2} e$. | ():2x-e,(n):-\frac{1}{2}x+\frac{3}{2}e | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,252 |
Example 2. Derive the equations of the tangent and normal to the curve $x^{2}+3 y^{2}=4$, drawn at the point with ordinate $y=1$ and negative abscissa. | Solution. When $y=1$, we find $x=-1$ and $x=1$. The problem mentions the point $A(-1,1)$. We find $y^{\prime}$ from the equation $x^{2}+3 y^{2}=4$. We have $2 x+6 y \cdot y^{\prime}=0$, hence $y^{\prime}=-\frac{x}{3 y}$, and thus, substituting $x=-1$, $y=1$, we get $k=y^{\prime}=\frac{1}{3}$.
Therefore, $(t): y=1+\fra... | ():\frac{1}{3}x+\frac{4}{3};\quad(n):-3x-2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,253 |
Example 3. Find the equations of the tangent and normal to the ellipse $x=4 \cos t, y=3 \sin t$, drawn at the point obtained when $t=\frac{\pi}{3}$. | Solution. We have: $x_{0}=4 \cos \frac{\pi}{3}=2, y_{0}=3 \sin \frac{\pi}{3}=3 \frac{\sqrt{3}}{2} ; y^{\prime}=$ $=\frac{d y}{d x}=\frac{3 \cos t}{-4 \sin t}=-\frac{3}{4} \operatorname{ctg} t ; k=-\frac{3}{4} \operatorname{ctg} \frac{\pi}{3}=-\frac{\sqrt{3}}{4}$. Therefore, $(t): y=\frac{3 \sqrt{3}}{2}-\frac{\sqrt{3}}{... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,254 |
Example 4. Investigate the monotonicity and extrema of the function $f(x)=\sqrt[3]{x^{2}}(x-1)$ | Solution. $f(x)$ is defined for all $x \in \mathbb{R}$. By the product rule formula, $f^{\prime}(x)=\frac{2}{3 \sqrt[3]{x}}(x-1)+\sqrt[3]{x^{2}}=\frac{5 x-2}{3 \sqrt[3]{x}} ; x=\frac{2}{5}$ and $x=0$ are the critical points of $f(x)$. On the upper axis of the diagram (Fig. 7.5), the distribution of the signs of $f^{\pr... | y_{\max}=0,y_{\}=-0.6\sqrt[3]{0.16} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,255 |
Example 5. Investigate the monotonicity and extrema of the function $y=2 x^{3}+3 x^{2}-12 x+1$. Also, find the maximum and minimum values of this function on the interval $[-1 ; 5]$. | Solution. We find \( y' \) and investigate its sign:
\[
y' = 6x^2 + 6x - 12 = 6(x-1)(x+2)
\]
The signs of \( y' \) and the intervals of monotonicity are shown in the diagram (Fig. 7.6); \( x = -2 \) and \( x = 1 \) are stationary points, \( x = 1 \) is a point of minimum, and \( x = -2 \) is a point of maximum. Here,... | y_{\}=-6,y_{\max}=21,y(5)=266 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,256 |
Example 6. Calculate the limits: a) $\lim _{x \rightarrow 1}\left(\frac{x}{x-1}-\frac{1}{\ln x}\right)$; b) $\lim _{x \rightarrow-\infty} x^{3} e^{x} ;$ c) $\lim _{x \rightarrow \pi / 2}(\cos x)^{(x-\pi / 2)} ; \quad$ d) $\lim _{x \rightarrow 0}(1-\sin x)^{\operatorname{ctg}^{2} x}$.
A visual overview indicates the pr... | Solution.
a) $\lim _{x \rightarrow 1}\left(\frac{x}{x-1}-\frac{1}{\ln x}\right)=(\infty-\infty)=\lim _{x \rightarrow 1} \frac{x \cdot \ln x-x+1}{(x-1) \ln x}=$
$$
\begin{aligned}
=\left(\frac{0}{0}\right) \stackrel{!}{=} \lim _{x \rightarrow 1} \frac{\ln x+1-1}{\ln x+\frac{x-1}{x}} & =\lim _{x \rightarrow 1} x \cdot \... | \frac{1}{2},0,1, | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,257 |
Example 1. $f(x)=\frac{x^{2}+5 x+4}{x^{2}-1}$. | Solution. $x=-1$ and $x=1-$ are points of discontinuity. We have:
$$
\begin{aligned}
\lim _{x \rightarrow-1} \frac{x^{2}+5 x+4}{x^{2}-1} & =\lim _{x \rightarrow-1} \frac{(x+1)(x+4)}{(x+1)(x-1)}=-\frac{3}{2} \\
\lim _{x \rightarrow 1-0} f(x) & =\lim _{x \rightarrow 1-0} \frac{x+4}{x-1}=-\infty \\
\lim _{x \rightarrow 1... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,258 |
Example 3. $\quad f(x)=e^{1 /\left[x^{2}(1-x)\right]}$. | Solution. $x=0$ and $x=1-$ are points of discontinuity,
\[
\begin{gathered}
\lim _{x \rightarrow 0} e^{1 /\left[x^{2}(1-x)\right]}=+\infty \\
\lim _{x \rightarrow 1-0} e^{1 /\left[x^{2}(1-x)\right]}=+\infty \\
\lim _{x \rightarrow 1+0} e^{1 /\left[x^{2}(1-x)\right]}=0
\end{gathered}
\]
Answer. $x=0, x=1$ - points of ... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,259 |
Example 4. $\quad f(x)=\frac{x^{3}+3 x^{2}}{x^{2}-2}$. | Solution.
$$
\begin{gathered}
k=\lim _{x \rightarrow \infty} \frac{f(x)}{x}=\lim _{x \rightarrow \infty} \frac{x^{3}+3 x^{2}-1}{x\left(x^{2}-2\right)}=1 \\
b=\lim _{x \rightarrow \infty}(f(x)-k x)=\lim _{x \rightarrow \infty} \frac{3 x^{2}+2 x-1}{x^{2}-2}=3
\end{gathered}
$$
$y=x+3-$ the oblique asymptote. | x+3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,260 |
Example 5. $f(x)=x+\sqrt{x}$. | Solution. The function is defined only for $x \geqslant 0$. There is no oblique asymptote, since
$$
k=\lim _{x \rightarrow \infty}\left(1+\frac{1}{\sqrt{x}}\right)=1, \quad b=\lim _{x \rightarrow+\infty} \sqrt{x}=+\infty
$$ | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,261 |
Example 6. $f(x)=\frac{\ln x}{\sqrt{x}}$. | Solution. The given function is defined for $x>0$, so $x \rightarrow \infty$ means $x \rightarrow+\infty$
$$
\begin{gathered}
k=\lim _{x \rightarrow \infty} \frac{\ln x}{x \sqrt{x}}=\left(\frac{\infty}{\infty}\right)=\lim _{x \rightarrow \infty} \frac{(\ln x)^{\prime}}{(x \sqrt{x})^{\prime}}=\left\{\begin{array}{c}
\t... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,262 |
Example 1. Investigate the convexity, concavity, and points of inflection of the function $y=x^{3}+3 x^{2}+6 x+7$. | Solution. We have: \( y' = 3x^2 + 6x + 6, y'' = 6x + 6 \). The signs of \( y'' \) are indicated on the number line.
Answer. In the interval \( (-\infty, -1) \), \( f(x) \) is convex, in the interval \( (-1, +\infty) \), \( f(x) \) is concave, \( x = -1 \) is the abscissa of the inflection point (Fig. 7.10) and \( y_{\... | y_{\infty}=3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,263 |
Example 2. Investigate the convexity, concavity, and points of inflection of the function $y=\sqrt[3]{x+2}$. We have: | Solution. It is necessary to investigate the sign of the second derivative.
$$
y^{\prime}=\frac{1}{3 \sqrt[3]{(x+2)^{2}}}, \quad y^{\prime \prime}=\frac{-2}{9 \sqrt[3]{(x+2)^{5}}}
$$
$y^{\prime \prime}$ is nowhere zero, but changes sign at the point $x=-2$. The signs of $y^{\prime \prime}$ are indicated on the diagra... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,264 |
Example 2. Expand the polynomial $y=x^{3}-2 x^{2}+3 x+5$ in terms of positive integer powers of the binomial $x-2$. | Solution. $y^{\prime}=3 x^{2}-4 x+3, y^{\prime \prime}=6 x-4, y^{\prime \prime \prime}=6, y^{I V}=0$. Next, $y(2)=11, y^{\prime}(2)=7, y^{\prime \prime}(2)=8, y^{\prime \prime \prime}(2)=6$. According to point $2^{\circ}$, we can write $y=11+7(x-2)+4(x-2)^{2}+(x-2)^{3}$. This is what was required.
Answer. $x^{3}-2 x^{... | x^{3}-2x^{2}+3x+5=11+7(x-2)+4(x-2)^{2}+(x-2)^{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,265 |
Example 3. Write the Taylor polynomial of the third degree centered at the point $x=3$ for the function $f(x)=\sqrt{1+x}$. | Solution. We have $T_{3}(x)=c_{0}+c_{1}(x-3)+c_{2}(x-3)^{2}+c_{3}(x-3)^{3}$, where $c_{k}=\left.\frac{f^{(k)}(x)}{k!}\right|_{x=3}$. First, let's find the derivatives:
$$
f^{\prime}(x)=\frac{1}{2 \sqrt{1+x}}, \quad f^{\prime \prime}(x)=\frac{-1}{4 \sqrt{(1+x)^{3}}}, \quad f^{\prime \prime \prime}(x)=\frac{3}{8 \sqrt{(... | T_{3}(x)=2+\frac{1}{4}(x-3)-\frac{1}{64}(x-3)^{2}+\frac{1}{512}(x-3)^{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,266 |
Example 1. Calculate the limits:
a) $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{\tan\left(x^{2}+y^{2}\right)}{x^{2}+y^{2}}$, b) $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x^{2}-y^{2}}{x^{2}+y^{2}}$, c) $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{x y}{\sqrt{4-x y}-2}$... | Solution. Note that the functions $\frac{\operatorname{tg}\left(x^{2}+y^{2}\right)}{x^{2}+y^{2}}$ and $\frac{x^{2}-y^{2}}{x^{2}+y^{2}}$ are undefined only at the point $(0,0)$, while $\frac{x y}{\sqrt{4-x y}-2}$ is undefined on the coordinate axes $x=0$ and $y=0$.
a) Transition to polar coordinates $x=r \cos \varphi, ... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,267 |
Example 1 (to $1^{\circ}$). Find the partial and total increments of the function $z=x^{2} y$ at the initial values $x=1, y=2$, if $\Delta x=0.1$; $\Delta y=-0.2$. | Solution. We have:
$$
\begin{aligned}
& \Delta_{x} z=(x+\Delta x)^{2} \cdot y-x^{2} y=(1+0.1)^{2} \cdot 2-1^{2} \cdot 2=0.42 \\
& \Delta_{y} z=x^{2}(y+\Delta y)-x^{2} y=1^{2} \cdot(2-0.2)-1^{2} \cdot 2=-0.2 \\
& \Delta z=(x+\Delta x)^{2} \cdot(y+\Delta y)-x^{2} y=(1+0.1)^{2} \cdot(2-0.2)-1^{2} \cdot 2=0.178 \\
& \Delt... | 0.178 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,268 |
Example 3 (to points $5^{\circ}, 6^{\circ}$ ). Find the partial differentials and the total differential of the function $z=x^{2} y^{3}$. Also find the values of these quantities at the point $(1,2)$. | Solution. We have: $z_{x}^{\prime}=2 x y^{3}, z_{y}^{\prime}=3 x^{2} y^{2}$. Therefore, $d_{x} z=2 x y^{3} d x, d_{y} z=3 x^{2} y^{2} d y, d z=2 x y^{3} d x+3 x^{2} y^{2} d y$. In particular, $d_{x} z(1,2)=16 d x, d_{y} z=12 d y, d z(1,2)=16 d x+12 d y-$ the differentials of independent variables coincide with their in... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,270 |
Example 1. Form the equations of the tangent plane and normal to the surface $z=2 x^{2}+y^{2}$ at the point $P_{0}(1,-1,3)$. | Solution. We have: $z_{x}^{\prime}=4 x, z_{y}^{\prime}=2 y ; z_{x}^{\prime}(1,-1)=4, z_{y}^{\prime}(1,-1)=-2$. Therefore,
$$
\begin{aligned}
& (t): z=3+4(x-1)-2(y+1), \quad \text { or } \quad 4 x-8 y-z-3=0 \\
& (n): \frac{x-1}{4}=\frac{y+1}{-2}=\frac{z-3}{-1}
\end{aligned}
$$ | \frac{x-1}{4}=\frac{y+1}{-2}=\frac{z-3}{-1} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,271 |
Example 2. Calculate approximately $1.08^{3.96}$. | Solution. Let $z=f(x, y)=x^{y}, x_{0}=1, y_{0}=4$; then $\Delta x=0.08, \Delta y=-0.04$.
We have: $z_{x}^{\prime}=y x^{y-1}, \quad f_{x}^{\prime}(1,4)=1, \quad z_{y}^{\prime}=x^{y} \ln x, \quad f_{y}^{\prime}(1,4)=0$, $z_{0}=f(1,4)=1$. Finally, according to the relation of point $4^{\circ}$, we get $1.08^{3.96} \appro... | 1.32 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,272 |
Example 3. Given the function $z=f(x, y)=x^{2}+y^{2}+2 x+1$ and two points $A\left(x_{0}, y_{0}\right)=A(2,3)$ and $B\left(x_{1}, y_{1}\right)=B(2.02,2.99)$. It is required:
1) to compute the value $z_{0}$ of the function $f(x, y)$ at point $A$ and the value $z_{1}$ at point $B$;
2) to compute the approximate value $\... | Solution. 1) We have $z_{1}(2.02,2.99)=(2.02)^{2}+(2.99)^{2}+2 \cdot 2.02+1=18.0605$
2) First, we find:
$$
\begin{gathered}
z_{0}=f\left(x_{0}, y_{0}\right)=2^{2}+3^{2}+2 \cdot 2+1=18 \\
f_{x}^{\prime}(x, y)=2 x+2, \quad f_{x}^{\prime}\left(x_{0}, y_{0}\right)=f^{\prime}(2,3)=2 \cdot 2+2=6 \\
f_{y}^{\prime}(x, y)=2 y... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,273 |
Example 1. Find $d^{2} z$, if $z=\frac{x+y}{x-y}$. | Solution. Sequential differentiation of the given function yields:
$$
\begin{gathered}
z_{x}^{\prime}=\frac{x-y-x-y}{(x-y)^{2}}=-\frac{2 y}{(x-y)^{2}}, \quad z_{y}^{\prime}=\frac{x-y+x+y}{(x-y)^{2}}=\frac{2 x}{(x-y)^{2}} \\
z_{x^{2}}^{\prime \prime}=\frac{4 y}{(x-y)^{3}}, \quad z_{x y}^{\prime \prime}=-2 \frac{x+y}{(x... | 4\frac{y^{2}-(x+y)y+z^{2}}{(x-y)^{3}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,274 |
Example 2. Find $d z, d^{2} z$ and $d^{3} z$, if $z=x^{5} y^{3}$. | Solution. We have:
$$
d z=d\left(x^{5} y^{3}\right)=5 x^{4} y^{3} d x+3 x^{5} y^{2} d y
$$
$d^{2} z=d\left(5 x^{4} y^{3} d x+3 x^{5} y^{2} d y\right)=20 x^{3} y^{3} d x^{2}+30 x^{4} y^{2} d x d y+6 x^{5} y d y^{2}$,
$d^{3} z=d\left(d^{2} z\right)=60 x^{2} y^{3} d x^{3}+180 x^{3} y^{2} d x^{2} d y+90 x^{4} y d x d y^... | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,275 |
Example 1. Find the derivative of the function $z=x^{2}+y^{2}$ at the point $M(3,1)$ in the direction of the point $M_{1}(0,5)$. | Solution. We have: $\overrightarrow{M M}_{1}=\{0-3,5-1\}=\{-3,4\},\left|\overrightarrow{M M}_{\mathrm{I}}\right|=$ $=\sqrt{(-3)^{2}+4^{2}}=5, \cos \alpha=-\frac{3}{5}, \sin \alpha=\frac{4}{5}$.
Let $\vec{l}=\left\{-\frac{3}{5}, \frac{4}{5}\right\}$ and find $\frac{\partial z}{\partial l}$.
We have: $z_{x}^{\prime}=2 ... | -2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,276 |
Example 2. Find the direction of the maximum increase of the function $z=3 x^{2}-2 y^{2}$ at the point $M(1,2)$. Also, find the greatest value of the derivatives in different directions at the point $M$. | Solution. Let's find the gradient of the function $z$ at the given point $(1,2)$. We have $z_{x}^{\prime}=6 x, z_{x}^{\prime}(1,2)=6, z_{y}^{\prime}=-4 y, z_{y}^{\prime}(1,2)=-8$. The gradient of the field at point $M(1,2)$ is $\overrightarrow{\operatorname{grad}} z=\{6,-8\}$. This vector indicates the direction of the... | 10 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,277 |
Example 3. Given the function $z=\arcsin \frac{y}{x^{2}}$, point $A(-2,-1)$, and vector $\vec{a}=3 \vec{i}-4 \vec{j}$. Find:
1) $\overrightarrow{\operatorname{grad}} z$ at point $A$;
2) the derivative at point $A$ in the direction of vector $\vec{a}$. | Solution. 1) To find the coordinates of the vector $\overrightarrow{\operatorname{grad}} z$ (see point $3^{\circ}$), we will find the partial derivatives
$$
z_{x}^{\prime}=\frac{1}{\sqrt{1-\left(\frac{y}{x^{2}}\right)^{2}}}\left(-\frac{y}{x^{4}}\right) \cdot 2 x=-\frac{2 y}{x \sqrt{x^{4}-y^{2}}} \quad \text { and } \q... | -\frac{1}{\sqrt{15}}\vec{i}+\frac{1}{\sqrt{15}}\vec{j} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,278 |
Example 1. Derive the Taylor formula for $n=2$ for the function $f(x, y)=x^{y}$ in the neighborhood of the point $M_{0}(1,1)$ and approximate $1.1^{1.02}$. | Solution. We have
$$
f(x, y)=f(1,1)+d f(1,1)+\frac{1}{2} d^{2} f(1,1)+R_{2}
$$
First, we find the partial derivatives of the first two orders:
$$
\begin{gathered}
f_{x}^{\prime}=y \cdot x^{y-1} ; \quad f_{y}^{\prime}=x^{y} \cdot \ln x \\
f_{x^{2}}^{\prime \prime}=y(y-1) x^{y-2} ; f_{x y}^{\prime \prime}=x^{y-1}+x^{y... | 1.102 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,279 |
Example 1 (to $1^{\circ}$). Investigate the function $z=$ $=x^{3}+y^{3}-3 x y$ for extremum. | Solution. The function is defined and differentiable for all $(x, y)$.
1) Find the stationary points:
$$
\left\{\begin{array} { l }
{ z _ { x } ^ { \prime } = 3 x ^ { 2 } - 3 y = 0 , } \\
{ z _ { y } ^ { \prime } = 3 y ^ { 2 } - 3 x = 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ x ^ { 2 } - y = 0 , ... | z_{\}=z(1,1)=-1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,280 |
Example 3 (to item $2^{\circ}$). On the hyperbola $x^{2}-y^{2}=9$, find the point that is closest to the point $A(0,-3)$. | Solution. Let's investigate the function expressing the square of the distance of point $M(x, y)$ from point $A(0,-3)$:
$$
A M^{2}=f(x, y)=x^{2}+(y+3)^{2}
$$
under the condition that the coordinates of point $M(x, y)$ satisfy the equation of the hyperbola $x^{2}-y^{2}-9=0$.
We form the Lagrange function:
$$
F(x, y)... | (-\frac{3\sqrt{5}}{2},-\frac{3}{2})(\frac{3\sqrt{5}}{2},-\frac{3}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,281 |
Example 1. Find the maximum and minimum values of the function $z=2 x^{3}-6 x y+3 y^{2}$ in the closed region bounded by the $O y$ axis, the line $y=2$, and the parabola $y=\frac{x^{2}}{2}$ for $x \geqslant 0$.
 The points at which the function takes its maximum and minimum values can be located both inside the area and on its boundary. We set the partial derivatives $\frac{\partial z}{\partial x}=6 x^{2}-6 y$ and $\frac{\partial z}{\partial y}=-6 x+6 y$ to zero.
Solvi... | z_{\text{max}}=z(0,2)=12,\quadz_{\text{}}=z(1,1)=-1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,282 |
Example 1. Form the equation of a straight line using the least squares method based on the data provided in the table.
| $x$ | 1 | 2 | 3 | 5 |
| :---: | :---: | :---: | :---: | :---: |
| $y$ | 3 | 4 | 2.5 | 0.5 | | Solution. Let's determine the coefficients of the system of equations:
$$
\begin{aligned}
& A_{1}=\sum x_{i}^{2}=39, B_{1}=\sum x_{i}=11, C_{1}=\sum x_{i} y_{i} \\
& A_{2}=11, B_{2}=4, C_{2}=\sum y_{i}=10
\end{aligned}
$$
From the system of equations
$$
\left\{\begin{array}{l}
39 k+11 p=21 \\
11 k+4 p=10
\end{array}... | -\frac{26}{35}x+\frac{159}{35} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,283 |
Example 1.1. Calculate
$$
\left|\begin{array}{rrr}
2 & -1 & 1 \\
3 & 2 & 2 \\
1 & -2 & 1
\end{array}\right|
$$ | Solution:
$$
\begin{gathered}
\left|\begin{array}{rrr}
2 & -1 & 1 \\
3 & 2 & 2 \\
1 & -2 & 1
\end{array}\right|=2 \cdot\left|\begin{array}{rr}
2 & 2 \\
-2 & 1
\end{array}\right|-(-1) \cdot\left|\begin{array}{ll}
3 & 2 \\
1 & 1
\end{array}\right|+1 \cdot\left|\begin{array}{rr}
3 & 2 \\
1 & -2
\end{array}\right|= \\
=2 ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,284 |
Example 1.3. Solve the system by matrix method
\[
\begin{aligned}
& 2 x_{1}-x_{2}+x_{3}=2 \\
& 3 x_{1}+2 x_{2}+2 x_{3}=-2 . \\
& x_{1}-2 x_{2}+x_{3}=1
\end{aligned}
\] | Solution. For the given system
$$
A=\begin{array}{rrr}
2 & -1 & 1 \\
3 & 2 & 2 \\
1 & -2 & 1
\end{array} ; X=\begin{gathered}
x_{1} \\
x_{2} \\
x_{3}
\end{gathered} ; B=-\begin{array}{r}
2 \\
1
\end{array}
$$
To use formula (1.9), we need to find the inverse matrix of matrix $A$. The example shows that the matrix $A^... | x_{1}=2;x_{2}=-1;x_{3}=-3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,285 |
Example 1.5. Solve the system of linear equations using the Gaussian method.
\[
\begin{aligned}
& 2 x_{1}+7 x_{2}+13 x_{3}=0 \\
& 3 x_{1}+14 x_{2}+12 x_{3}=18 \\
& 5 x_{1}+25 x_{2}+16 x_{3}=39
\end{aligned}
\] | Solution. Let's take the first equation of the system as the first leading equation, and the first leading unknown as $x_{1}$; the first leading element will be $a_{11}=2$. We will eliminate $x_{1}$ from the second and third equations by adding the leading equation, multiplied by $-\frac{3}{2}$, to the second equation,... | x_{1}=-4,x_{2}=3,x_{3}=-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,286 |
Example 1.6. Solve the system of linear equations
$$
\begin{aligned}
& 2 x_{1}+5 x_{2}-4 x_{3}=8 \\
& 3 x_{1}+15 x_{2}-9 x_{3}=5 \\
& 5 x_{1}+5 x_{2}-7 x_{3}=1
\end{aligned}
$$ | Solution. We will transform the system using the Gaussian method:
$$
\begin{array}{ll}
2 x_{1}+5 x_{2}-4 x_{3}=8, & 2 x_{1}+5 x_{2}-4 x_{3}=8 \\
\frac{15}{2} x_{2}-3 x_{3}=-7, & \frac{15}{2} x_{2}-3 x_{3}=-7, \\
-\frac{15}{2} x_{2}+3 x_{3}=-19, & 0=-26
\end{array}
$$
The equation $0=-26$ makes no sense, hence the giv... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,287 |
Example 1.7. Solve the system of linear equations
$$
\begin{aligned}
& 2 x_{1}+5 x_{2}-4 x_{3}=8 \\
& 3 x_{1}+15 x_{2}-9 x_{3}=5 \\
& 5 x_{1}+5 x_{2}-7 x_{3}=27
\end{aligned}
$$ | Solution. We transform the system using the Gaussian method:
$$
\begin{array}{ll}
2 x_{1}+5 x_{2}-4 x_{3}=8, & 2 x_{1}+5 x_{2}-4 x_{3}=8, \\
\frac{15}{2} x_{2}-3 x_{3}=-7, & \frac{15}{2} x_{2}-3 x_{3}=-7 \\
-\frac{15}{2} x_{2}+3 x_{3}=7 ; &
\end{array}
$$
After the second step, out of three equations, only two remain... | \begin{aligned}&x_{1}=\frac{19}{3}+x_{3}\\&x_{2}=-\frac{14}{15}+\frac{2}{5}x_{3}\end{aligned} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,288 |
Example 1.16. Given the vertices $A(2 ; 1), B(6 ; 3), C(4 ; 5)$ of a triangle. Find: 1) the length of side $A B ; 2)$ the internal angle $A$ in radians with an accuracy of 0.01; 3) the equation of the altitude passing through vertex $C ; 4$ ) the equation of the median passing through vertex $C$; 5) the point of inters... | Solution.
We make a drawing (Fig. 16).
Fig. 16
1. The length of side $AB$ is found as the distance between two points $A$ and $B$.
$$
\begin{gathered}
|AB|=\sqrt{\left(x_{B}-x_{A}\right... | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,289 | |
Example 1.17. Form the equation of the line $l$ passing through the point $A(2; -4)$ and being at a distance of 2 units from the origin. | Solution. Let the equation of the desired line have the form:
$$
y-y_{A}=k\left(x-x_{A}\right),
$$
or
$$
y+4=k(x-2)
$$
or
$$
k x-y-(4+2 k)=0 .
$$
To determine the slope $k$ of this line, we will use the fact that it is 2 units away from the origin. We will find this distance directly. The equation of the perpendi... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,290 |
Example 1.18. Form the equations of lines parallel to the line $3 x+4 y-1=0(l)$ and at a distance of 1 from it. | Solution. We will seek the equation of each line in the form $y=k x+b$. Since the desired line is parallel to the line $l$, its slope $k=-\frac{3}{4}$, and therefore, its equation takes the form:
$$
y=-\frac{3}{4} x+b
$$
or
$$
3 x+4 y-4 b=0
$$
To find the parameter $b$, we will use the fact that the distance from a... | 3x+4y-6=0\quad\text{}\quad3x+4y+4=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,291 |
Example 1.21. Form the equation of the plane passing through the line of intersection of the planes $x+y+5z-1=0$, $2x+3y-z+2=0$ and through the point $M(3,2,1)$. | Solution. We will use the equation of a pencil of planes
$$
x+y+5 z-1+\lambda(2 x+3 y-z+2)=0 .
$$
The value of $\lambda$ is determined from the condition that the coordinates of point $M$ must satisfy this equation:
$$
\begin{gathered}
3+2+5-1+\lambda(6+6-1+2)=9+13 \lambda=0 \\
\lambda=-\frac{9}{13}
\end{gathered}
$... | 5x+14y-74z+31=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,292 |
Example 1.22. Form the equation of the plane passing through the line of intersection of the planes $x+3 y+5 z-4=0$ and $x-y-2 z+7=0$ and parallel to the y-axis. | Solution. We will use the equation of the pencil $x+3 y+5 z-4+$ $+\lambda(x-y-2 z+7)=0, \quad$ transform the equation to the form $(1+\lambda) x+(3-\lambda) y+(5-2 \lambda) z+(7 \lambda-4)=0$.
Since the desired plane is parallel to the y-axis, the coefficient of y should be zero, i.e., $3-\lambda=0, \lambda=3$. Substi... | 4x-z+17=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,293 |
Example 1.23. Find the equation of the plane passing through the points $M(2, -1, 4)$ and $N(3, 2, -1)$ and perpendicular to the plane $x+y+z-3=0$. | Solution. We will use the equation of the plane passing through the first of the given points:
$$
A(x-2)+B(y+1)+C(z-4)=0
$$
The condition for this plane to pass through the second point and the condition of perpendicularity are given by the equations:
$$
\begin{gathered}
A+3 B-5 C=0 \\
A+B+C=0
\end{gathered}
$$
By ... | 4x-3y-z-7=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,294 |
Example 1.24. Perpendiculars are dropped from the point $P(2 ; 3 ;-5)$ to the coordinate planes. Find the equation of the plane passing through their feet. | Solution. The bases of the perpendiculars dropped onto the coordinate planes will be the following points $M_{1}(2 ; 3 ; 0), M_{2}(2 ; 0 ;-5)$, $M_{3}(0 ; 3 ;-5)$. To write the equation of the plane passing through points $M_{1}, M_{2}, M_{3}$, we will use the equation
$$
\left|\begin{array}{rrr}
x-x_{1} & y-y_{1} & z... | 15x+10y-6z-60=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,295 |
Example 1.25. Form the equation of the plane passing through the point $M(2 ; 3 ; 5)$ and perpendicular to the vector $\bar{N}=4 \bar{i}+3 \bar{j}+2 \bar{k}$ | Solution. It is sufficient to use the equation of the plane passing through the given point and perpendicular to the given vector:
$$
4(x-2)+3(y-3)+2(z-5)=0, \quad \text { i.e. } \quad 4 x+3 y+2 z-27=0 .
$$
## Line.
1. A line can be defined by the equations of two - intersecting planes
$$
\begin{aligned}
& A_{1} x+... | 4x+3y+2z-27=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,296 |
Example 1.26. Reduce to canonical form the equations of the line $2 x-y+3 z-1=0$ and $5 x+4 y-z-7=0$. | Solution. Excluding $y$ first, and then $z$, we get:
$$
13 x+11 z-11=0 \quad \text { and } \quad 17 x+11 y-22=0 .
$$
If we solve each of the equations for $x$, we will have:
$$
x=\frac{11(y-2)}{-17}=\frac{11(z-1)}{-13}
$$
from which
$$
\frac{x}{-11}=\frac{y-2}{17}=\frac{z-1}{13} \text {. }
$$
Second method: find ... | \frac{x}{-11}=\frac{y-2}{17}=\frac{z-1}{13} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,297 |
Example 1.27. Construct the line
\[
\begin{array}{r}
2 x+3 y+3 z-9=0 \\
4 x+2 y+z-8=0
\end{array}
\] | Solution. The desired line can be constructed as the line of intersection of planes. For this, we write the equations of the planes that define the line in intercept form on the axes:
$$
\begin{aligned}
& \frac{x}{4.5}+\frac{y}{3}+\frac{z}{3}=1 \\
& \frac{x}{2}+\frac{y}{4}+\frac{z}{8}=1
\end{aligned}
$$
By constructi... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,298 |
Example 1.28. Drop a perpendicular from the origin to the line
$$
\frac{x-2}{2}=\frac{y-1}{3}=\frac{z-3}{1} \text {. }
$$ | Solution. Let's form the equation of the plane passing through the origin and perpendicular to the given line: $2 x+3 y+z=0$. (For this plane, we can take $A=l ; B=m$; $C=n ; D=0 ;$ the condition of perpendicularity of the line and the plane is used, see point 8 of the introduction to this section).
Let's find the poi... | \frac{x}{1}=\frac{y}{-2}=\frac{z}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,299 |
Example 1.29. In the equations of the line $\frac{x}{2}=\frac{y}{-3}=\frac{z}{n}$, determine the parameter $n$ so that this line intersects with the line $\frac{x+1}{3}=\frac{y+5}{2}=\frac{z}{1}$, and find the point of their intersection. | Solution. To find the parameter $n$, we use the condition of the intersection of 2 straight lines:
$$
\left|\begin{array}{ccc}
x_{1}-x_{0} & y_{1}-y_{0} & z_{1}-z_{0} \\
l_{1} & m_{1} & n_{1} \\
l_{2} & m_{2} & n_{2}
\end{array}\right|=0
$$
From this, it follows:
$$
\left|\begin{array}{rrr}
1 & 5 & 0 \\
3 & 2 & 1 \\... | M(2,-3,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,300 |
Example 1.30. A line is given by the canonical equations
$$
\frac{x-2}{3}=\frac{y+1}{5}=\frac{z-3}{-1} \text {. }
$$
Form the general equations of this line. | Solution. The canonical equations of the line can be written as a system of two independent equations:
$$
\begin{aligned}
& \frac{x-2}{3}=\frac{y+1}{5} \\
& \frac{x-2}{3}=\frac{z-3}{-1}, \text { from which } \quad \begin{array}{r}
5 x-3 y-13=0 \\
x+3 z-11=0
\end{array}
\end{aligned}
$$
We have obtained the general eq... | \begin{aligned}&5x-3y-13=0\\&x+3z-11=0\end{aligned} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,301 |
Example 1.31. Find the coordinates of the point $M$ that bisects the segment of the line
$$
\frac{x-2}{3}=\frac{y+1}{5}=\frac{z-3}{-1}
$$
enclosed between the planes $x o z$ and $x o y$. | Solution. To find the point $A$ of intersection of the line with the $x o z$ plane, we set $y=0$ in the equations of the line. Then we get:
$$
\frac{x-2}{3}=\frac{1}{5}=\frac{z-3}{-1}, \text { or } \quad \begin{aligned}
& \frac{x-2}{3}=\frac{1}{5} \\
& \frac{z-3}{-1}=\frac{1}{5}
\end{aligned}
$$
from which $\quad x=2... | M(6.8;7;1.4) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,302 |
Example 1.32. Form the equation of the plane passing through the line
\[
\begin{aligned}
3 x+2 y+5 z+6 & =0 \\
x+4 y+3 z+4 & =0
\end{aligned}
\]
parallel to the line
\[
\frac{x-1}{3}=\frac{y-5}{2}=\frac{z+1}{-3}
\] | Solution. Let's form the equation of the pencil of planes passing through the first of the given lines:
$$
\alpha(3 x+2 y+5 z+6)+\beta(x+4 y+3 z+4)=0
$$
which we divide by $\alpha \neq 0$, and let $\beta / \alpha=\lambda$ :
$$
\begin{aligned}
& 3 x+2 y+5 z+6+\lambda(x+4 y+3 z+4)=0, \text { or } \\
& (3+\lambda) x+(2... | 2x+3y+4z+5=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,303 |
Example 1.33. Given a line
\[
\begin{aligned}
3 x-2 y-z+4 & =0 \\
x-4 y-3 z-2 & =0
\end{aligned}
\]
Find its projection on the plane
\[
5 x+2 y+2 z-7=0 .
\] | Solution. We need to find a plane that passes through the given line and is perpendicular to the given plane; then the desired projection will be determined as the intersection of this plane with the given one.
Let's form the equation of the pencil of planes passing through the given line:
$$
\begin{aligned}
& (3 x-2... | \frac{x-1}{-2}=\frac{y-1}{-14}=\frac{z}{19} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,304 |
Example 2.26. Find the intervals of monotonicity of the function $y=x^{4}-\frac{4}{3} x^{3}+5$. | Solution. The domain of the given function is the entire number line. Differentiating, we find
$$
y^{\prime}=4 x^{3}-4 x^{2}=4 x^{2}(x-1) .
$$
The derivative $y^{\prime}$ has no points of discontinuity. The zeros of the derivative $y^{\prime}$ are the roots of the equation $x^{2}(x-1)=0$, i.e., $x_{1}=0$ and $x_{2}=1... | (-\infty;1) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,305 |
Example 2.27. Find the extrema of the function $y=x \sqrt[33]{(x-1)^{2}}$. | Solution. The function is defined on the entire number line. Let's compute the derivative:
$$
y^{\prime}=3 x^{2} \sqrt[3]{(x-1)^{2}}+x^{3} \cdot \frac{2}{3 \sqrt[3]{x-1}}=\frac{x^{2}}{3 \sqrt[3]{x-1}}(9 x-9+2 x)=\frac{x^{2}(11 x-9)}{3 \sqrt[3]{x-1}}
$$
The derivative $f^{\prime}$ is zero at $x=0$ and $x=\frac{9}{11}$... | y(\frac{9}{11})=\frac{729}{1331}\sqrt[3]{\frac{4}{121}}\approx0.176 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,306 |
Example 2.28. Find the intervals of convexity and concavity and the points of inflection of the graph of the function $f(x)=3 x^{5}+5 x^{4}-20 x^{3}+60 x-5$. | Solution. The function is defined on the entire number line. Differentiating it twice, we get
$$
\begin{gathered}
f^{\prime}(x)=15 x^{4}+20 x^{3}-60 x^{2}+60 \\
f^{\prime \prime}(x)=60 x^{3}+60 x^{2}-120 x=60 x(x-1)(x+2)
\end{gathered}
$$
The second derivative exists on the entire number line and is zero at $x=-2, x=... | (-2;19),(0;-5),(1;43) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,307 |
Example 2.29. Find the maximum and minimum values of the following functions:
1) $f(x)=x^{3}-3 x^{2}+4$ on the interval $[1,3]$
2) $\phi(x)=x+\frac{1}{x} \quad$ on the interval $[-2,2]$. | Solution.
1. The function $f(x)$ is continuous on the interval $[1,3]$. We find
$$
f^{\prime}(x)=3 x^{2}-6 x
$$
In this case, the critical points are only those where the second derivative $f^{\prime \prime}(x)$ is zero, i.e., $x=0$ and $x=2$. Only one of these critical points, $x=2$, belongs to the interval $[1,3]$... | 40forf(x);nomaximumorminimumfor\phi(x) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,308 |
Example 2.30. Plot the graph of the function $y=x^{5}-x^{3}-2 x$. | ## Solution.
1. The function is defined and continuous on the interval ( $-\infty ; \infty$ ).
2. The function is odd, i.e., $y(-x) = -y(x)$, which means
$$
y(-x) = (-x)^{5} - (-x)^{3} - 2(-x) = -\left(x^{5} - x^{3} - 2 x\right) = -y(x)
$$
Therefore, the graph of the function is symmetric with respect to the origin,... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,309 |
2.1.1. A ray given by the equation $l_{1}: 3 x+4 y-18=0$ falls on the line $l: 3 x+2 y-12=0$, which is capable of reflecting rays. Form the equation of the reflected ray. | Solution. Since the angle of incidence of the ray is equal to the angle of reflection of the ray, then $\angle \phi_{1}=\angle \phi_{2}$, i.e., $\operatorname{tg} \phi_{1}=\operatorname{tg} \phi_{2}$ (Fig. 31).
$ to the distance from the same point to the line $l: x=-4$ is 2. Reduce the equation of the line to its canonical form and determine the type of this curve. | Solution. Let $M(x ; y)$ be the current point on the line. From point $M$, we drop a perpendicular to the line $x=-4$, which intersects it at point $N(-4 ; y)$. According to the problem's condition: $\frac{|M F|}{|M N|}=2$, or $|M F|=2|M N|$.
Then
$$
\sqrt{\left(x_{M}-x_{F}\right)^{2}+\left(y_{M}-y_{F}\right)^{2}}=2 ... | \frac{X^{2}}{4^{2}}-\frac{Y^{2}}{(4\sqrt{3})^{2}}=1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 30,311 |
3.2.1. Find the limits of functions:
a) $\lim _{x \rightarrow \pm \infty} \sqrt{x^{2}-2 x-1}-\sqrt{x^{2}-7 x+3}$ (indeterminate form $\infty-\infty$ ). | Solution. Multiply and divide the given expression by $\sqrt{x^{2}-2 x-1}+\sqrt{x^{2}-7 x+3}$:
$$
\begin{aligned}
& \sqrt{x^{2}-2 x-1}-\sqrt{x^{2}-7 x+3} \quad \sqrt{x^{2}-2 x-1}+\sqrt{x^{2}-7 x+3} \\
& \lim _{x \rightarrow \pm \infty} \sqrt{\sqrt{x^{2}-2 x-1}+\sqrt{x^{2}-7 x+3}}= \\
& =\lim _{x \rightarrow \pm \infty... | \frac{5}{2},24\sqrt{2},2,e^{-2},-\frac{1}{8},e^{-\frac{1}{8}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,312 |
3.4.1. Find the equations of the tangents to the curve $y=\frac{3 x+2}{3 x-2}$ that are parallel to the line $3 x+y+3=0$. | Solution. Let $M\left(x_{0} ; y_{0}\right)$ be the coordinates of the point of tangency of the tangent and the curve. We choose the equation of the tangent in the form:
$$
y-y_{0}=k_{k a c .} \cdot\left(x-x_{0}\right)
$$
Since the tangent is parallel to the line $3 x+y+3=0$, the slope of which is $k_{n p .}=-3$, then... | 3x+y+1=0\text{}3x+y-7=0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,313 |
3.4.2. Find the greatest and the least values of the function $f(x)=2 x^{3}+3 x^{2}-36 x+1$ on the interval $[1 ; 11]$. | Solution. The given function is continuous on $[1 ; 11]$. We find $f^{\prime}(x)=6 x^{2}+6 x-36=0$. Setting $y^{\prime}=0$, we have $x^{2}+x-6=0$,
from which we obtain two critical points: $x_{1}=-3$ and $x_{2}=2$. To find the minimum and maximum values of the function on the given interval, it is sufficient to compute... | -432630 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,314 |
### 3.1.1. Antiderivative Function and Indefinite Integral
In differential calculus, given a function $F(x)$, one finds its derivative $f(x)=F^{\prime}(x)$. In practice, it is often necessary to solve the inverse problem: to restore the function $F(x)$, knowing its derivative $f(x)$. In this case, the function $F(x)$ ... | Solution. $\int 2 x d x=x^{2}+C ; 1^{2}+C=3 ; C=2$.
The desired antiderivative: $F(x)=x^{2}+2$.
Theorem 2 (on the existence of an indefinite integral).
If the function $f(x)$ is continuous on the interval $[a, b]$, then an indefinite integral of this function exists on the same interval. | F(x)=x^{2}+2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,315 |
Example 3.4. Find the indefinite integral
$$
\int \frac{(x+1) d x}{\sqrt{3-x^{2}}}
$$
and verify the result by differentiation. | Solution. We decompose the given integral into the sum of two integrals:
$$
\int \frac{(x+1) d x}{\sqrt{3-x^{2}}}=\int \frac{x d x}{\sqrt{3-x^{2}}}+\int \frac{d x}{\sqrt{3-x^{2}}}
$$
To compute the first of these integrals, we use the fact that
$$
x d x=\frac{1}{2} d\left(x^{2}\right)=-\frac{1}{2} d\left(3-x^{2}\rig... | -\sqrt{3-x^{2}}+\arcsin\frac{x}{\sqrt{3}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,316 |
Example 3.10. Find the integral $\int \frac{d x}{\sqrt{2 x+3}+\sqrt[3]{(2 x+3)^{2}}}$. | Solution. Note that $\sqrt{2 x+3}=(2 x+3)^{\frac{1}{2}}, \sqrt[3]{(2 x+3)^{2}}=(2 x+3)^{\frac{2}{3}}$. The least common multiple of the denominators of the fractions $\frac{1}{2}, \frac{2}{3}$ is 6. Therefore, if we apply the substitution $2 x+3=t^{6}$, we will have:
$$
t=(2 x+3)^{\frac{1}{6}}, \sqrt{2 x+3}=\left(t^{6... | \frac{3}{2}\sqrt[3]{2x+3}-3\sqrt[6]{2x+3}+3\ln|\sqrt[6]{2x+3}+1|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,317 |
Example 3.13. Using the Newton-Leibniz formula, compute the definite integral
$$
\int_{3}^{8} \frac{x}{\sqrt{1+x}} d x
$$ | Solution. To get rid of the irrationality, we make the substitution $1+x=t^{2}$. Then
$$
\begin{aligned}
& \int_{3}^{8} \frac{x d x}{\sqrt{1+x}}=\left|\begin{array}{l}
1+x=t^{2} \\
d x=2 t d t \\
x=3 \rightarrow t=2 \\
x=8 \rightarrow t=3
\end{array}\right|=\int_{2}^{3} \frac{\left(t^{2}-1\right) 2 t d t}{t}=2 \int_{2... | \frac{32}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,318 |
Example 3.18. Calculate the volume of the body formed by rotating the figure bounded by one half-wave of the sine curve $y=\sin x$ around the $O x$ axis. | ## Solution.
$$
\begin{gathered}
V=\pi \int_{0}^{\pi} y^{2} d x=\pi \int_{0}^{\pi} \sin ^{2} x d x=\pi \int_{0}^{\pi} \frac{1-\cos 2 x}{2} d x=\frac{\pi}{2} \int_{0}^{\pi}(1-\cos 2 x) d x= \\
=\frac{\pi}{2} x-\left.\frac{1}{2} \sin 2 x\right|_{0} ^{\pi}=\frac{\pi^{2}}{2} .
\end{gathered}
$$ | \frac{\pi^{2}}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,319 |
Example 3.19. Calculate the volume of the body obtained by rotating around the Ox axis the figure bounded by the parabolas $y=x^{2}$ and $x=y^{2}$ (Fig. 43).
Fig. 43 | Solution. Solving the system of equations
$$
\begin{aligned}
& y=x^{2} \\
& y^{2}=x, x^{4}=x, x^{4}-x=0, x(x-1)\left(x^{2}+x+1\right)=0
\end{aligned}
$$
we get $x_{1}=0, x_{2}=1, y_{1}=0, y_{2}=1$, from which the points of intersection of the curves are $O(0 ; 0), B(1 ; 1)$. As can be seen (Fig. 43), the volume of th... | \frac{3}{10}\pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 30,320 |
Example 3.20. Find the domain of the function
$$
z=\frac{1}{\sqrt{4-x^{2}-y^{2}}}
$$ | Solution. The function has real values if $4-x^{2}-$ $-y^{2}>0$ or $x^{2}+y^{2}<4$. The latter inequality is satisfied by the coordinates of points lying inside a circle of radius 2 centered at the origin. The domain of existence of the function is the set of points inside this circle (Fig. 44).
![](https://cdn.mathpi... | x^{2}+y^{2}<4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 30,321 |
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