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Use Gaussian elimination to solve the given \( 2 \times 2 \) system of equations.\n\n\[ \n{2x} + y = 1 \n\]\n\n\[ \n{4x} + {2y} = 6 \n\] | Solution Write the system as an augmented matrix.\n\n\[ \n\left\lbrack \begin{array}{lll} 2 & 1 & 1 \\ 4 & 2 & 6 \end{array}\right\rbrack \n\]\n\nObtain a 1 in row 1, column 1 . This can be accomplished by multiplying the first row by \( \frac{1}{2} \) .\n\n\[ \n\frac{1}{2}{R}_{1} = {R}_{1} \rightarrow \left\lbrack \be... | Yes |
Solve the system of equations.\n\n\\[ \n{3x} + {4y} = {12} \n\\]\n\n\\[ \n{6x} + {8y} = {24} \n\\] | Solution Perform row operations on the augmented matrix to try and achieve row-echelon form.\n\n\\[ \nA = \\left\\lbrack \\begin{array}{lll} 3 & 4 & {12} \\\\ 6 & 8 & {24} \\end{array}\\right\\rbrack \n\\]\n\n\\[ \n- \\frac{1}{2}{R}_{2} + {R}_{1} = {R}_{1} \\rightarrow \\left\\lbrack \\begin{matrix} 0 & 0 & 0 \\\\ 6 & ... | Yes |
Perform row operations on the given matrix to obtain row-echelon form. | \[ \left\lbrack \begin{array}{rrrr} 1 & - 3 & 4 & 3 \\ 2 & - 5 & 6 & 6 \\ - 3 & 3 & 4 & 6 \end{array}\right\rbrack \] \n\nSolution The first row already has a 1 in row 1, column 1. The next step is to multiply row 1 by -2 and add it to row 2. \n\nThen replace row 2 with the result. \n\n\[ - 2{R}_{1} + {R}_{2} = {R}_{2}... | Yes |
Show that the given matrices are multiplicative inverses of each other. | Solution Multiply \( {AB} \) and \( {BA} \) . If both products equal the identity, then the two matrices are inverses of each other.\n\n\[{AB} = \left\lbrack \begin{array}{rr} 1 & 5 \\ - 2 & - 9 \end{array}\right\rbrack \left\lbrack \begin{array}{rr} - 9 & - 5 \\ 2 & 1 \end{array}\right\rbrack\n\]\n\n\[= \left\lbrack \... | Yes |
Use the formula to find the multiplicative inverse of matrix \( A \) | Solution Using the formula, we have\n\n\[ {A}^{-1} = \frac{1}{\left( 1\right) \left( {-3}\right) - \left( {-2}\right) \left( 2\right) }\left\lbrack \begin{array}{ll} - 3 & 2 \\ - 2 & 1 \end{array}\right\rbrack \]\n\n\[ = \frac{1}{-3 + 4}\left\lbrack \begin{array}{ll} - 3 & 2 \\ - 2 & 1 \end{array}\right\rbrack \]\n\n\[... | Yes |
Find the inverse, if it exists, of the given matrix.\n\[ A = \left\lbrack \begin{array}{ll} 3 & 6 \\ 1 & 2 \end{array}\right\rbrack \] | Solution We will use the method of augmenting with the identity.\n\n\[ \left\lbrack \begin{array}{llll} 3 & 6 & 1 & 0 \\ 1 & 3 & 0 & 1 \end{array}\right\rbrack \]\n\n1. Switch row 1 and row 2.\n\n\[ \left\lbrack \begin{array}{llll} 1 & 3 & 0 & 1 \\ 3 & 6 & 1 & 0 \end{array}\right\rbrack \]\n\n2. Multiply row 1 by -3 an... | Yes |
Find the determinant of the \( 3 \times 3 \) matrix given\n\n\[ A = \left\lbrack \begin{array}{rrr} 0 & 2 & 1 \\ 3 & - 1 & 1 \\ 4 & 0 & 1 \end{array}\right\rbrack \] | Solution Augment the matrix with the first two columns and then follow the formula. Thus,\n\n\[ \left| A\right| = \left| \begin{array}{rrrrr} 0 & 2 & 1 & 0 & 2 \\ 3 & - 1 & 1 & 3 & - 1 \\ 4 & 0 & 1 & 4 & 2 \end{array}\right| \]\n\n\[ = 0\left( {-1}\right) \left( 1\right) + 2\left( 1\right) \left( 4\right) + 1\left( 3\r... | Yes |
Find the solution to the given \( 3 \times 3 \) system using Cramer’s Rule.\n\n\[ x + y - z = 6 \]\n\n\[ {3x} - {2y} + z = - 5 \]\n\n\[ x + {3y} - {2z} = {14} \] | Solution Use Cramer's Rule.\n\n\[ D = \left| \begin{array}{rrr} 1 & 1 & - 1 \\ 3 & - 2 & 1 \\ 1 & 3 & - 2 \end{array}\right| ,{D}_{x} = \left| \begin{array}{rrr} 6 & 1 & - 1 \\ - 5 & - 2 & 1 \\ {14} & 3 & - 2 \end{array}\right| ,{D}_{y} = \left| \begin{array}{rrr} 1 & 6 & - 1 \\ 3 & - 5 & 1 \\ 1 & {14} & - 2 \end{array... | Yes |
Solve the system of equations using Cramer's Rule. | \[ \n{3x} - {2y} = 4 \n\] \n \n\[{6x} - {4y} = 0\] \n \nSolution We begin by finding the determinants \( D,{D}_{x} \), and \( {D}_{y} \) . \n \n\[D = \left| \begin{array}{ll} 3 & - 2 \\ 6 & - 4 \end{array}\right| = 3\left( {-4}\right) - 6\left( {-2}\right) = 0\] \n \nWe know that a determinant of zero means that either... | Yes |
Graph the ellipse given by the equation, \( \frac{{\left( x + 2\right) }^{2}}{4} + \frac{{\left( y - 5\right) }^{2}}{9} = 1 \) . Identify and label the center, vertices, co-vertices, and foci. | Solution First, we determine the position of the major axis. Because \( 9 > 4 \), the major axis is parallel to the \( y \) -axis.\n\nTherefore, the equation is in the form \( \frac{{\left( x - h\right) }^{2}}{{b}^{2}} + \frac{{\left( y - k\right) }^{2}}{{a}^{2}} = 1 \), where \( {b}^{2} = 4 \) and \( {a}^{2} = 9 \) . ... | Yes |
Graph \( {y}^{2} = {24x} \) . Identify and label the focus, directrix, and endpoints of the latus rectum. | Solution The standard form that applies to the given equation is \( {y}^{2} = {4px} \) . Thus, the axis of symmetry is the \( x \) -axis. It follows that:\n\n- \( {24} = {4p} \), so \( p = 6 \) . Since \( p > 0 \), the parabola opens right\n\n- the coordinates of the focus are \( \left( {p,0}\right) = \left( {6,0}\righ... | Yes |
Graph \( {x}^{2} = - {6y} \) . Identify and label the focus, directrix, and endpoints of the latus rectum. | Solution The standard form that applies to the given equation is \( {x}^{2} = {4py} \) . Thus, the axis of symmetry is the \( y \) -axis.\n\nIt follows that:\n\n- \( - 6 = {4p} \), so \( p = - \frac{3}{2} \) Since \( p < 0 \), the parabola opens down.\n\n- the coordinates of the focus are \( \left( {0, p}\right) = \lef... | Yes |
Graph \( {\left( y - 1\right) }^{2} = - {16}\left( {x + 3}\right) \). Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum. | Solution The standard form that applies to the given equation is \( {\left( y - k\right) }^{2} = {4p}\left( {x - h}\right) \). Thus, the axis of symmetry is parallel to the \( x \) -axis. It follows that:\n\n- the vertex is \( \left( {h, k}\right) = \left( {-3,1}\right) \)\n\n- the axis of symmetry is \( y = k = 1 \)\n... | Yes |
Graph \( r = \frac{10}{5 - 4\cos \theta } \) . | Solution First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 5, which is \( \frac{1}{5} \) .\n\n\[ r = \frac{10}{5 - 4\cos \theta } = \frac{{10}\left( \frac{1}{5}\right) }{5\left( \frac{1}{5}\right) - 4\left( \frac{1}{5}\right) \cos \theta } \]\n\n\[ r = \frac{... | Yes |
Find the polar form of the conic given a focus at the origin, \( e = 3 \) and directrix \( y = - 2 \) . | The directrix is \( y = - p \), so we know the trigonometric function in the denominator is sine.\n\nBecause \( y = - 2, - 2 < 0 \), so we know there is a subtraction sign in the denominator. We use the standard form of\n\n\[ r = \frac{ep}{1 - e\sin \theta } \]\n\nand \( e = 3 \) and \( \left| {-2}\right| = 2 = p \) .\... | Yes |
Find the polar form of a conic given a focus at the origin, \( e = \frac{3}{5} \), and directrix \( x = 4 \) . | Solution Because the directrix is \( x = p \), we know the function in the denominator is cosine. Because \( x = 4,4 > 0 \), so we know there is an addition sign in the denominator. We use the standard form of\n\n\[ r = \frac{ep}{1 + e\cos \theta } \]\n\nand \( e = \frac{3}{5} \) and \( \left| 4\right| = 4 = p \) .\n\n... | Yes |
Write the first five terms of the sequence. | Solution Substitute \( n = 1, n = 2 \), and so on in the formula.\n\n\[ n = 1\;{a}_{1} = \frac{{\left( -1\right) }^{1}{1}^{2}}{1 + 1} = - \frac{1}{2} \]\n\n\[ n = 2\;{a}_{2} = \frac{{\left( -1\right) }^{2}{2}^{2}}{2 + 1} = \frac{4}{3} \]\n\n\[ n = 3\;{a}_{3} = \frac{{\left( -1\right) }^{3}{3}^{2}}{3 + 1} = - \frac{9}{4... | Yes |
Write the first five terms of the arithmetic sequence with \( {a}_{1} = {17} \) and \( d = - 3 \) . | Solution Adding -3 is the same as subtracting 3. Beginning with the first term, subtract 3 from each term to find the next term.\n\nThe first five terms are \( \{ {17},{14},{11},8,5\} \) | Yes |
Given \( {a}_{1} = 8 \) and \( {a}_{4} = {14} \), find \( {a}_{5} \). | Solution The sequence can be written in terms of the initial term 8 and the common difference \( d \).\n\n\[ \{ 8,8 + d,8 + {2d},8 + {3d}\} \]\n\nWe know the fourth term equals 14; we know the fourth term has the form \( {a}_{1} + {3d} = 8 + {3d} \).\n\nWe can find the common difference \( d \).\n\n\[ {a}_{n} = {a}_{1}... | Yes |
Write an explicit formula for the arithmetic sequence.\n\n$ \\{ 2,{12},{22},{32},{42},\\ldots \\} $ | Solution The common difference can be found by subtracting the first term from the second term.\n\n\[ d = {a}_{2} - {a}_{1} \]\n\n\[ = {12} - 2 \]\n\n\[ = {10} \]\n\nThe common difference is 10 . Substitute the common difference and the first term of the sequence into the formula and simplify.\n\n\[ {a}_{n} = 2 + {10}\... | Yes |
Find the number of terms in the finite arithmetic sequence.\n\n\[ \n\{ 8,1, - 6,\ldots , - {41}\} \n\] | Solution The common difference can be found by subtracting the first term from the second term.\n\n\[ \n1 - 8 = - 7 \n\]\n\nThe common difference is -7. Substitute the common difference and the initial term of the sequence into the \( n \) th term formula and simplify.\n\n\[ \n{a}_{n} = {a}_{1} + d\left( {n - 1}\right)... | Yes |
Is the sequence geometric? If so, find the common ratio. | Solution Divide each term by the previous term to determine whether a common ratio exists.\n\na. \( \frac{2}{1} = 2\;\frac{4}{2} = 2\;\frac{8}{4} = 2\;\frac{16}{8} = 2 \)\n\nThe sequence is geometric because there is a common ratio. The common ratio is 2 .\n\nb. \( \frac{12}{48} = \frac{1}{4}\;\frac{4}{12} = \frac{1}{3... | Yes |
Evaluate \( \mathop{\sum }\limits_{{k = 3}}^{7}{k}^{2} \) . | Solution According to the notation, the lower limit of summation is 3 and the upper limit is 7 . So we need to find the sum of \( {k}^{2} \) from \( k = 3 \) to \( k = 7 \) . We find the terms of the series by substituting \( k = 3,4,5,6 \), and 7 into the function \( {k}^{2} \) . We add the terms to find the sum.\n\n\... | Yes |
Find the sum of each arithmetic series.\na. \( 5 + 8 + {11} + {14} + {17} + {20} + {23} + {26} + {29} + {32} \) | a. We are given \( {a}_{1} = 5 \) and \( {a}_{n} = {32} \).\nCount the number of terms in the sequence to find \( n = {10} \).\nSubstitute values for \( {a}_{1},{a}_{n} \), and \( n \) into the formula and simplify.\n\[ {S}_{n} = \frac{n\left( {{a}_{1} + {a}_{n}}\right) }{2} \]\n\[ {S}_{10} = \frac{{10}\left( {5 + {32}... | Yes |
Example 4 Finding the First \( n \) Terms of a Geometric Series\n\nUse the formula to find the indicated partial sum of each geometric series.\n\n a. \( {S}_{11} \) for the series \( 8 + \left( {-4}\right) + 2 + \ldots \)\n\n b. \( \mathop{\sum }\limits_{{k = 1}}^{6}3 \cdot {2}^{k} \) | Solution\n\n a. \( {a}_{1} = 8 \), and we are given that \( n = {11} \).\n\n We can find \( r \) by dividing the second term of the series by the first.\n\n \[ r = \frac{-4}{8} = - \frac{1}{2} \]\n\n Substitute values for \( {a}_{1}, r \), and \( n \) into the formula and simplify.\n\n \[ {S}_{n} = \frac{{a}_{1}\left( ... | Yes |
Find each binomial coefficient.\n\na. \( \\left( \\begin{array}{l} 5 \\ 3 \\end{array}\\right) \) b. \( \\left( \\begin{array}{l} 9 \\ 2 \\end{array}\\right) \) c. \( \\left( \\begin{array}{l} 9 \\ 7 \\end{array}\\right) \) | Use the formula to calculate each binomial coefficient. You can also use the \( {}_{n}{C}_{r} \) function on your calculator.\n\n\[ \left( \\begin{array}{l} n \\ r \\end{array}\\right) = C\\left( {n, r}\\right) = \\frac{n!}{r!\\left( {n - r}\\right) !} \]\n\n\[ \\text{a.}\\left( \\begin{array}{l} 5 \\ 3 \\end{array}\\r... | Yes |
Evaluate \( \mathop{\lim }\limits_{{x \rightarrow 3}}\left( {{2x} + 5}\right) \) . | \[ \mathop{\lim }\limits_{{x \rightarrow 3}}\left( {{2x} + 5}\right) = \mathop{\lim }\limits_{{x \rightarrow 3}}\left( {2x}\right) + \mathop{\lim }\limits_{{x \rightarrow 3}}\left( 5\right) \]\nSum of functions property\n\[ = 2\mathop{\lim }\limits_{{x \rightarrow 3}}\left( x\right) + \mathop{\lim }\limits_{{x \rightar... | Yes |
Evaluate \( \mathop{\lim }\limits_{{x \rightarrow 3}}\left( {5{x}^{2}}\right) \) . | \[ \mathop{\lim }\limits_{{x \rightarrow 3}}\left( {5{x}^{2}}\right) = 5\mathop{\lim }\limits_{{x \rightarrow 3}}\left( {x}^{2}\right) \] Constant times a function property \[ = 5\left( {3}^{2}\right) \] Function raised to an exponent property \[ = {45} \] | Yes |
Evaluate \( \mathop{\lim }\limits_{{x \rightarrow 5}}\left( {2{x}^{3} - {3x} + 1}\right) \) . | \[ \mathop{\lim }\limits_{{x \rightarrow 5}}\left( {2{x}^{3} - {3x} + 1}\right) = \mathop{\lim }\limits_{{x \rightarrow 5}}\left( {2{x}^{3}}\right) - \mathop{\lim }\limits_{{x \rightarrow 5}}\left( {3x}\right) + \mathop{\lim }\limits_{{x \rightarrow 5}}\left( 1\right) \]\nSum of functions\n\[ = 2\mathop{\lim }\limits_{... | Yes |
Evaluate \( \mathop{\lim }\limits_{{x \rightarrow 2}}{\left( 3x + 1\right) }^{5} \) . | Solution We will take the limit of the function as \( x \) approaches 2 and raise the result to the \( {5}^{\text{th }} \) power.\n\n\[ \mathop{\lim }\limits_{{x \rightarrow 2}}{\left( 3x + 1\right) }^{5} = {\left( \mathop{\lim }\limits_{{x \rightarrow 2}}\left( 3x + 1\right) \right) }^{5} \]\n\n\[ = {\left( 3\left( 2\... | Yes |
Evaluate \( \mathop{\lim }\limits_{{x \rightarrow 2}}\left( \frac{{x}^{2} - {6x} + 8}{x - 2}\right) \) . | \[ \mathop{\lim }\limits_{{x \rightarrow 2}}\left( \frac{{x}^{2} - {6x} + 8}{x - 2}\right) = \mathop{\lim }\limits_{{x \rightarrow 2}}\left( \frac{\left( {x - 2}\right) \left( {x - 4}\right) }{x - 2}\right) \;\text{Factor the numerator.} \] \[ = \mathop{\lim }\limits_{{x \rightarrow 2}}\left( \frac{\left( {x - 2}\right... | Yes |
Evaluate \( \mathop{\lim }\limits_{{x \rightarrow 5}}\left( \frac{\frac{1}{x} - \frac{1}{5}}{x - 5}\right) \) . | Solution Find the LCD for the denominators of the two terms in the numerator, and convert both fractions to have\nthe LCD as their denominator.\n\n\[ \mathop{\lim }\limits_{{x \rightarrow 5}}\left( \frac{\frac{1}{x} - \frac{1}{5}}{x - 5}\right) = \mathop{\lim }\limits_{{x \rightarrow 5}}\left( \frac{{5x}\left( {\frac{1... | Yes |
Evaluate \( \mathop{\lim }\limits_{{x \rightarrow 0}}\left( \frac{\sqrt{{25} - x} - 5}{x}\right) \) . | \[ \mathop{\lim }\limits_{{x \rightarrow 0}}\left( \frac{\sqrt{{25} - x} - 5}{x}\right) = \mathop{\lim }\limits_{{x \rightarrow 0}}\left( {\frac{\sqrt{{25} - x} - 5}{x} \cdot \frac{\sqrt{{25} - x} + 5}{\sqrt{{25} - x} + 5}}\right) \] Multiply numerator and denominator by the conjugate. \[ = \mathop{\lim }\limits_{{x \r... | Yes |
Evaluate \( \mathop{\lim }\limits_{{x \rightarrow 4}}\left( \frac{4 - x}{\sqrt{x} - 2}\right) \) . | \[ \mathop{\lim }\limits_{{x \rightarrow 4}}\left( \frac{4 - x}{\sqrt{x} - 2}\right) = \mathop{\lim }\limits_{{x \rightarrow 4}}\left( \frac{\left( {2 + \sqrt{x}}\right) \left( {2 - \sqrt{x}}\right) }{\sqrt{x} - 2}\right) \] Factor. \[ = \mathop{\lim }\limits_{{x \rightarrow 4}}\left( \frac{\left( {2 + \sqrt{x}}\right)... | Yes |
Evaluate \( \mathop{\lim }\limits_{{x \rightarrow 7}}\frac{\left| x - 7\right| }{x - 7} \) . | The function is undefined at \( x = 7 \), so we will try values close to 7 from the left and the right.\n\nLeft-hand limit: \( \frac{\left| {6.9} - 7\right| }{{6.9} - 7} = \frac{\left| {6.99} - 7\right| }{{6.99} - 7} = \frac{\left| {6.999} - 7\right| }{{6.999} - 7} = - 1 \)\n\nRight-hand limit: \( \frac{\left| {7.1} - ... | Yes |
Determine whether the function \( f\left( x\right) = \frac{{x}^{2} - {25}}{x - 5} \) is continuous at \( x = 5 \) . | Solution To determine if the function \( f \) is continuous at \( x = 5 \), we will determine if the three conditions of continuity are satisfied at \( x = 5 \) . Condition 1:\n\n\( f\left( 5\right) \) does not exist.\n\n\( \Rightarrow \) Condition 1 fails.\n\nThere is no need to proceed further. Condition 2 fails at \... | No |
Example 6 Estimating the Derivative at a Point on the Graph of a Function\n\nFrom the graph of the function \( y = f\\left( x\\right) \) presented in Figure 5, estimate each of the following:\n\n\[ f\\left( 0\\right) ;f\\left( 2\\right) ;{f}^{\\prime }\\left( 0\\right) ;{f}^{\\prime }\\left( 2\\right) \] | Solution To find the functional value, \( f\\left( a\\right) \), find the \( y \) -coordinate at \( x = a \) .\n\nTo find the derivative at \( x = a,{f}^{\\prime }\\left( a\\right) \), draw a tangent line at \( x = a \), and estimate the slope of that tangent line. See Figure 6.\n\n![f7473d67-b7fd-4fb3-af38-0b0c856cf58... | Yes |
Theorem 2.6.1. If \( f\left( x\right) \) is not defined for \( x = a \), and if\n\n\[ \mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) = B \]\n\nthen \( f\left( x\right) \) will be continuous for \( x = a \), if \( B \) is assumed as the value of \( f\left( x\right) \) for \( x = a \) . | Thus the function\n\n\[ \frac{{x}^{2} - 4}{x - 2} \]\n\nis not defined for \( x = 2 \) (since then there would be division by zero). But for every other value of \( x \) ,\n\n\[ \frac{{x}^{2} - 4}{x + 2} = x + 2 \]\n\nand\n\n\[ \mathop{\lim }\limits_{{x \rightarrow 2}}\left( {x + 2}\right) = 4 \]\n\ntherefore \( \matho... | No |
Theorem 2.8.2. The limit of the product of a finite number of variables is equal to the product of the limits of the several variables. | \[ \mathop{\lim }\limits_{{x \rightarrow a}}\left\lbrack {f\left( x\right) \cdot g\left( x\right) }\right\rbrack = \mathop{\lim }\limits_{{x \rightarrow a}}f\left( x\right) \cdot \mathop{\lim }\limits_{{x \rightarrow a}}g\left( x\right) . \] | Yes |
Example 3.2.1. For instance, consider the function\n\n\\[ \ny = {x}^{2} \n\\]\n\nAssuming \\( x = {10} \\) for the initial value of \\( x \\) fixes \\( y = {100} \\) as the initial value of \\( y \\) . Suppose \\( x \\) increases to \\( x = {12} \\), that is, \\( {\\Delta x} = 2 \\) ; then \\( y \\) increases to \\( y ... | Sage\n\nsage: \\( x = \\operatorname{var}\\left( {\\text{\ | No |
Differentiate \( 3{x}^{2} + 5 \) . | Applying the successive steps in the General Rule, we get, after placing\n\n\[ y = 3{x}^{2} + 5 \]\n\nFirst step.\n\n\[ y + {\Delta y} = 3\left( {x + {\Delta x}}\right) 2 + 5 = 3{x}^{2} + {6x} \cdot {\Delta x} + 3{\left( \Delta x\right) }^{2} + 5. \]\n\nSecond step.\n\n\[ y + {\Delta y} = 3{x}^{2} + {6x} \cdot {\Delta ... | Yes |
Differentiate \( {x}^{3} - {2x} + 7 \) . | Solution. Place \( y = {x}^{3} - {2x} + 7 \) .\n\nFirst step.\n\n\[ y + {\Delta y} = \left( {x + {\Delta x}}\right) 3 - 2\left( {x + {\Delta x}}\right) + 7 \]\n\n\[ = {x}^{3} + 3{x}^{2} \cdot {\Delta x} + {3x} \cdot {\left( \Delta x\right) }^{2} + {\left( \Delta x\right) }^{3} - {2x} - 2 \cdot {\Delta x} + 7 \]\n\nSeco... | Yes |
Differentiate \( \frac{c}{{x}^{2}} \) . | Place \( y = \frac{c}{{x}^{2}} \) . \n\nFirst step. \( y + {\Delta y} = \frac{c}{{\left( x + \Delta x\right) }^{2}} \) . \n\nSecond step. \n\n\[ \ny + {\Delta y} = \frac{c}{{\left( x + \Delta x\right) }^{2}} \n\] \n\n\[ \ny = \frac{c}{{x}^{2}} \n\] \n\n\[ \n{\Delta y}\; = \frac{c}{{\left( x + \Delta x\right) }^{2}} - \... | Yes |
Differentiate \( f\left( x\right) \), where\n\n\[ f\left( x\right) = \left\{ \begin{array}{ll} - 2{x}^{3} + 3{x}^{2}, & 0 < x < 1, \\ 0, & x \leq 0\text{ or }x \geq 1. \end{array}\right. \] | \nThe function is given in parts, so the problem must be solved case-by-case. First, assume \( 0 < x < 1 \) .\n\n\( 0 < x < 1 \) : In this case, the derivative can be computed ass in the examples above to show\n\n\[ {f}^{\prime }\left( x\right) = - 6{x}^{2} + {6x},\;0 < x < 1. \]\n\nThis is not the final answer though!... | No |
Find the slopes of the tangents to the parabola \( y = {x}^{2} \) at the vertex, and at the point where \( x = \frac{1}{2} \) . | Solution. Differentiating by General Rule, (§3.7), we get\n\n\[ \n{y}^{\prime } = \frac{dy}{dx} = {2x} = \text{ slope of tangent line at any point on curve. } \n\] \n\nTo find slope of tangent at vertex, substitute \( x = 0 \) in \( {y}^{\prime } = {2x} \), giving \n\n\[ \n\frac{dy}{dx} = 0 \n\] \n\nTherefore the tange... | Yes |
If \( f\left( x\right) = {x}^{2} \), for \( x > 0 \), and \( \phi \left( y\right) = \sqrt{y} \), then the graphs are Now flip this graph about the \( {45}^{ \circ } \) line: | Let us now differentiate the inverse functions\n\n\[ y = f\left( x\right) \text{ and }x = \phi \left( y\right) \]\n\nsimultaneously by the General Rule.\n\n- FIRST STEP. \( y + {\Delta y} = f\left( {x + {\Delta x}}\right), x + {\Delta x} = \phi \left( {y + {\Delta y}}\right) \)\n\n- SECOND STEP.\n\n\[ y + {\Delta y} = ... | No |
Differentiate \( y = \log \left( {{x}^{2} + a}\right) \) . | By (4.11) (with \( v = {x}^{2} + a \) ), we have\n\n\[ \frac{dy}{dx} = \frac{\frac{d}{dx}\left( {{x}^{2} + a}\right) }{{x}^{2} + a} \]\n\n\[ = \frac{2x}{{x}^{2} + a}\text{. } \] | Yes |
Differentiate \( y = \log \sqrt{1 - {x}^{2}} \) . | By (4.11) and (4.6),\n\n\[ \frac{dy}{dx} = \frac{\frac{d}{dx}{\left( 1 - {x}^{2}\right) }^{\frac{1}{2}}}{{\left( 1 - {x}^{2}\right) }^{\frac{1}{2}}} \]\n\n\[ = \frac{\frac{1}{2}{\left( 1 - {x}^{2}\right) }^{-\frac{1}{2}}\left( {-{2x}}\right) }{{\left( 1 - {x}^{2}\right) }^{\frac{1}{2}}} \]\n\n\[ = \frac{x}{{x}^{2} - 1}... | Yes |
Differentiate \( y = {a}^{3{x}^{2}} \) . | By (4.12),\n\n\[ \frac{dy}{dx} = \log a \cdot {a}^{3{x}^{2}}\frac{d}{dx}\left( {3{x}^{2}}\right) \]\n\n\[ = {6x}\log a \cdot {a}^{3{x}^{2}}. \] | Yes |
Differentiate \( y = b{e}^{{c}^{2} + {x}^{2}} \) . | By (4.4) and (4.13),\n\n\[ \frac{dy}{dx} = b\frac{d}{dx}\left( {e}^{{c}^{2} + {x}^{2}}\right) \]\n\n\[ = b{e}^{{c}^{2} + {x}^{2}}\frac{d}{dx}\left( {{c}^{2} + {x}^{2}}\right) \]\n\n\[ = {2bx}{e}^{{c}^{2} + {x}^{2}}\text{.} \] | Yes |
Differentiate \( y = {x}^{{e}^{x}} \) . | By (4.14),\n\n\[ \frac{dy}{dx} = {e}^{x}{x}^{{e}^{x} - 1}\frac{d}{dx}\left( x\right) + {x}^{{e}^{x}}\log x\frac{d}{dx}\left( {e}^{x}\right) \]\n\n\[ = {e}^{x}{x}^{{e}^{x} - 1} + {x}^{{e}^{x}}\log x \cdot {e}^{x} \]\n\n\[ = {e}^{x}{x}^{{e}^{x}}\left( {\frac{1}{x} + \log x}\right) \] | Yes |
Differentiate \( y = \log \sqrt{1 - {x}^{2}} \) . | By using 10, in §12.1, we may write this in a form free from radicals as follows: \( y = \frac{1}{2}\log \left( {1 - {x}^{2}}\right) \) . Then\n\n\[ \frac{dy}{dx} = \frac{1}{2}\frac{\frac{d}{dx}\left( {1 - {x}^{2}}\right) }{1 - {x}^{2}}\text{by (4.11)} \]\n\n\[ = \frac{\overline{1}}{2} \cdot \frac{-\overline{2}}{1 - {x... | Yes |
Differentiate \( y = \log \sqrt{\frac{1 + {x}^{2}}{1 - {x}^{2}}} \) . | Solution. Simplifying by means of (10) and (8), in §12.1,\n\n\[ y = \frac{1}{2}\left\lbrack {\log \left( {1 + {x}^{2}}\right) - \log \left( {1 - {x}^{2}}\right) }\right\rbrack \]\n\n\[ \frac{dy}{dx} = \frac{1}{2}\left\lbrack {\frac{\frac{d}{dx}\left( {1 + {x}^{2}}\right) }{1 + {x}^{2}} - \frac{\frac{d}{dx}\left( {1 - {... | Yes |
Differentiate \( y = {x}^{{e}^{x}} \) . | Taking the logarithm of both sides, \( \log y = {e}^{x}\log x \), by Formula 9 in [12.1]. Now differentiate both sides with respect to \( x \) :\n\n\[ \frac{dy}{dx} = {e}^{x}\frac{d}{dx}\left( {\log x}\right) + \log x\frac{d}{dx}\left( {e}^{x}\right) \text{by (4.10) and (4.5)} \]\n\n\[ = {e}^{x} \cdot \frac{1}{x} + \lo... | Yes |
Differentiate \( y = {\left( 4{x}^{2} - 7\right) }^{2 + \sqrt{{x}^{2} - 5}} \) . | Solution. Taking the logarithm of both sides,\n\n\[ \log y = \left( {2 + \sqrt{{x}^{2} - 5}}\right) \log \left( {4{x}^{2} - 7}\right) .\n\]\n\nDifferentiating both sides with respect to \( x \) ,\n\n\[ \frac{1}{y}\frac{dy}{dx} = \left( {2 + \sqrt{{x}^{2} - 5}}\right) \frac{8x}{4{x}^{2} - 7} + \log \left( {4{x}^{2} - 7}... | Yes |
Differentiate \( y = \sqrt{\frac{\left( {x - 1}\right) \left( {x - 2}\right) }{\left( {x - 3}\right) \left( {x - 4}\right) }} \) . | Solution. Taking the logarithm of both sides,\n\n\[ \log y = \frac{1}{2}\left\lbrack {\log \left( {x - 1}\right) + \log \left( {x - 2}\right) - \log \left( {x - 3}\right) - \log \left( {x - 4}\right) }\right\rbrack . \]\n\nDifferentiating both sides with respect to \( x \) ,\n\n\[ \frac{1}{y}\frac{dy}{dx} = \frac{1}{2}... | Yes |
Show that the hikers are climbing downhill when \( x = 5 \) . | This can be verified \ | No |
Given the curve \( y = \frac{{x}^{3}}{3} - {x}^{2} + 2 \) (see Figure 5.3).\n\n(a) Find \( \tau \) when \( x = 1 \) . | (a) \( \tan \tau = {\left\lbrack \frac{dy}{dx}\right\rbrack }_{x = 1}^{\infty } = 1 - 2 = - 1 \) ; therefore \( \tau = {135}^{ \circ } = {3\pi }/4 \) . | Yes |
Find the angle of intersection of the circles\n\n(A) \( {x}^{2} + {y}^{2} - {4x} = 1 \) ,\n\n(B) \( {x}^{2} + {y}^{2} - {2y} = 9 \) . | Solution. Solving simultaneously, we find the points of intersection to be \( \left( {3,2}\right) \) and \( \left( {1, - 2}\right) \) . This can be verified \ | No |
For example, take equation of circle\n\n\[ \n{x}^{2} + {y}^{2} = {r}^{2}\\text{ or }y = \\sqrt{{r}^{2} - {x}^{2}}. \n\] | We have\n\n\[ \n\\left\\{ \\begin{array}{l} x = r\\cos t \\\\ y = r\\sin t \\end{array}\\right. \n\]\n\n(5.9)\n\nas parametric equations of the circle, \\( t \\) being the parameter \\( {}^{5} \\) .\n\nIf we eliminate \\( t \\) between equations (5.9) by squaring and adding the results, we have\n\n\[ \n{x}^{2} + {y}^{2... | Yes |
Newtonian physics tells us that\n\n\[ \left\{ \begin{array}{l} x = {v}_{0}\cos \alpha \cdot t, \\ y = - \frac{1}{2}g{t}^{2} + {v}_{0}\sin \alpha \cdot t \end{array}\right. \]\n\nare really the parametric equations of the trajectory of a projectile \( 6 \), the time \( t \) being the parameter. The elimination of \( t \... | Since from (5.8) \( y \) is given as a function of \( t \), and \( t \) as a function of \( x \), we have\n\n\[ \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}\text{by (4.27)} \]\n\n\[ = \frac{dy}{dt} \cdot \frac{1}{\frac{dx}{dt}}\;\text{by (4.28)} \]\n\nthat is,\n\n\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{d... | Yes |
Find equations of tangent and normal, lengths of subtangent and subnormal to the ellipse\n\n\[ \left\{ \begin{array}{l} x = a\cos \phi \\ y = b\sin \phi \end{array}\right. \]\n\n(5.11)\n\nat the point where \( \phi = \frac{\pi }{4} \) . | Solution. The parameter being \( \phi ,\frac{dx}{d\phi } = - a\sin \phi ,\frac{dy}{d\phi } = b\cos \phi \) .\n\nSubstituting \( \phi = \frac{\pi }{4} \) in the given equations (5.11), we get \( \left( {\frac{a}{\sqrt{2}},\frac{b}{\sqrt{2}}}\right) \) as the point of contact. Hence \( {\left\lbrack \frac{dy}{dx}\right\r... | Yes |
Given equation of the cycloid in parametric form\n\n\[ \left\{ \begin{array}{l} x = a\left( {\theta - \sin \theta }\right) \\ y = a\left( {1 - \cos \theta }\right) \end{array}\right. \]\n\n\( \theta \) being the variable parameter; find lengths of subtangent, subnormal, tangent, and normal at the point where \( \theta ... | Solution:\n\n\[ \frac{dx}{d\theta } = a\left( {1 - \cos \theta }\right) ,\;\frac{dy}{d\theta } = a\sin \theta . \]\n\nSubstituting in (5.10),\n\n\[ \frac{dy}{dx} = \frac{\sin \theta }{1 - \cos \theta } \]\n\nthe slope at any point. Since \( \theta = \frac{\pi }{2} \), the point of contact is \( \left( {\frac{\pi a}{2} ... | Yes |
Find \( \psi \) and \( \tau \) in the cardioid \( \psi = a\left( {1 - \cos \theta }\right) \). Also find the slope at \( \theta = \frac{\pi }{6} \) . | Solution. \( \frac{d\widetilde{\psi }}{d\theta } = a\sin \theta \) . Substituting in (5.12) gives\n\n\[ \tan \psi = \frac{\rho }{\frac{d\rho }{d\theta }} = \frac{a\left( {1 - \cos \theta }\right) }{a\sin \theta } = \frac{{2a}{\sin }^{2}\frac{\theta }{2}}{{2a}\sin \frac{\theta }{2}\cos \frac{\theta }{2}} = \tan \frac{\t... | Yes |
Find the angle of of intersection of the curves \( \rho = a\sin {2\theta } \) , \( \rho = a\cos {2\theta }. \) | Solving the two equations simultaneously, we get at the point of intersection\n\n\[ \n\tan {2\theta } = 1,\;{2\theta } = {45}^{ \circ } = \pi /4,\;\theta = \frac{{45}^{ \circ }}{2} = \pi /8. \n\]\n\nFrom the first curve, using (5.12),\n\n\[ \n\tan {\psi }^{\prime } = \frac{1}{2}\tan {2\theta } = \frac{1}{2} \n\]\n\nfor... | Yes |
Find lengths of polar subtangent and subnormal to the lemniscate \( {\rho }^{2} = {a}^{2}\cos {2\theta } \) | Solution. Differentiating the equation of the curve as an implicit function with respect to \( \theta \), or, \( {2\rho }\frac{d\rho }{d\theta } = - 2{a}^{2}\sin {2\theta },\frac{d\rho }{d\theta } = - \frac{{a}^{2}\sin {2\theta }}{\rho } \). Substituting in (5.16) and (5.17), we get \[ \text{length of polar subtangent}... | Yes |
Example 5.10.1. Solve the equation \( {x}^{3} - 8{x}^{2} + {13x} - 6 = 0 \) . | Solution. Place \( f\left( x\right) = {x}^{3} - 8{x}^{2} + {13x} - 6 \) . \n\nFirst step. \( {f}^{\prime }\left( x\right) = 3{x}^{2} - {16x} + {13} \) . \n\nSecond step. G.C.D. \( = x - 1 \) . \n\nThird step. \( x - 1 = 0 \), therefore \( x = 1 \) . \n\nSince 1 occurs once as a root in the G.C.D., it will occur twice i... | Yes |
Consider the problem of finding the positive number \( x \) with \( \cos \left( x\right) = {x}^{3} \). | We can rephrase that as finding the zero of \( f\left( x\right) = \cos \left( x\right) - {x}^{3} \) . We have \( {f}^{\prime }\left( x\right) = - \sin \left( x\right) - 3{x}^{2} \) . Since \( \cos \left( x\right) \leq 1 \) for all \( x \) and \( {x}^{3} > 1 \) for \( x > 1 \) , we know that our zero lies between 0 and ... | Yes |
Given \( y = {e}^{ax} \), find \( \frac{{d}^{n}y}{d{x}^{n}} \). | Solution. \( \frac{dy}{dx} = a{e}^{ax},\frac{{d}^{2}y}{d{x}^{2}} = {a}^{2}{e}^{ax},\ldots ,\frac{{d}^{n}y}{d{x}^{n}} = {a}^{n}{e}^{ax} \). | Yes |
Given \( y = \log x \), find \( \frac{{d}^{n}y}{d{x}^{n}} \). | Solution. \( \frac{dy}{dx} = \frac{1}{x},\frac{{d}^{2}y}{d{x}^{2}} = - \frac{1}{{x}^{2}},\frac{{d}^{3}y}{d{x}^{3}} = \frac{1.2}{{x}^{3}},\frac{{d}^{4}y}{d{x}^{4}} = \frac{1 \cdot 2 \cdot 3}{{x}^{4}},\ldots \frac{{d}^{n}y}{d{x}^{n}} = {\left( -1\right) }^{n - 1}\frac{\left( {n - 1}\right) !}{{x}^{n}} \). | Yes |
Given \( y = \sin x \), find \( \frac{{d}^{n}y}{d{x}^{n}} \). | Solution. \( \frac{dy}{dx} = \cos x = \sin \left( {x + \frac{\pi }{2}}\right) \), \[ \frac{{d}^{2}y}{d{x}^{2}} = \frac{d}{dx}\sin \left( {x + \frac{\pi }{2}}\right) = \cos \left( {x + \frac{\pi }{2}}\right) = \sin \left( {x + \frac{2\pi }{2}}\right) , \] \[ \frac{{d}^{3}y}{d{x}^{3}} = \frac{d}{dx}\sin \left( {x + \frac... | Yes |
Given \( y = {e}^{x}\log x \), find \( \frac{{d}^{3}y}{d{x}^{3}} \) by Leibnitz’s Formula. | Let \( u = {e}^{x} \), and \( v = \log x \) ; then \( \frac{du}{dx} = {e}^{x},\frac{dv}{dx} = \frac{1}{x},\frac{{d}^{2}u}{d{x}^{2}} = {e}^{x} \) , \( \frac{{d}^{2}v}{d{x}^{2}} = - \frac{1}{{x}^{2}},\frac{{d}^{3}u}{d{x}^{3}} = {e}^{x},\frac{{d}^{3}v}{d{x}^{3}} = \frac{2}{{x}^{3}} \).\n\nSubstituting in (6.1), we get\n\n... | Yes |
Given \( y = {x}^{2}{e}^{ax} \), find \( \frac{{d}^{n}y}{d{x}^{n}} \) by Leibnitz’s Formula. | Solution. Let \( u = {x}^{2} \), and \( v = {e}^{ax} \) ; then \( \frac{du}{dx} = {2x},\frac{dv}{dx} = a{e}^{ax},\frac{{d}^{2}u}{d{x}^{2}} = {2x} \) , \( \frac{{d}^{2}v}{d{x}^{2}} = {a}^{2}{e}^{ax},\frac{{d}^{3}u}{d{x}^{3}} = 0,\frac{{d}^{3}v}{d{x}^{3}} = {a}^{3}{e}^{ax},\ldots ,\frac{{d}^{n}u}{d{x}^{n}} = 0,\frac{{d}^... | Yes |
A wooden box is to be built to contain \( {108}{\mathrm{{ft}}}^{3} \) . It is to have an open top and a square base. What must be its dimensions in order that the amount of material required shall be a minimum; that is, what dimensions will make the cost the least? | Let \( x \) denote the length of side of square base in feet, and \( y \) denote the height of box. Since the volume of the box is given, \( y \) may be found in terms of \( x \) . Thus volume \( = {x}^{2}y = {108} \), so \( y = \frac{108}{{x}^{2}} \) . Let \( M \) denote the number of square feet of lumber required as... | Yes |
Consider the function \( y = {a}^{x}, a > 1 \), whose graph is shown in Figure 7.5. | As we move along the curve from left to right the curve is rising; that is, as \( x \) increases the function \( y = {a}^{x} \) always increases. Therefore \( {a}^{x}\left( {a > 1}\right) \) is an increasing function for all values of \( x \) . | Yes |
In the problem worked out in Example 7.1.2, we showed by means of the graph of the function\n\n\\[ \nA = x\\sqrt{{100} - {x}^{2}} \n\\]\n\nthat the rectangle of maximum area inscribed in a circle of radius 5 inches contained 50 square inches. This may now be proved analytically as follows by applying the above rule. | Solution. Let \\( f\\left( x\\right) = x\\sqrt{{100} - {x}^{2}} \\) .\n\nFirst step. Compute \\( {f}^{\\prime }\\left( x\\right) = \\frac{{100} - 2{x}^{2}}{\\sqrt{{100} - {x}^{2}}} \\) .\n\nSecond step. \\( \\frac{{100} - 2{x}^{2}}{\\sqrt{{100} - {x}^{2}}} = 0 \\) implies \\( x = 5\\sqrt{2} \\), which is the critical v... | Yes |
Let us now apply the above rule to test analytically the function\n\n\[ M = {x}^{2} + \frac{432}{x} \] | Solution. Let \( f\left( x\right) = {x}^{2} + \frac{432}{x} \) .\n\nFirst step. Compute \( {f}^{\prime }\left( x\right) = {2x} - \frac{432}{{x}^{2}} \) .\n\nSecond step. Solve \( {2x} - \frac{432}{{x}^{2}} = 0 \) . (In Example 7.1.1 we got \( x = 6 \) .)\n\nThird step. Compute \( {f}^{\prime \prime }\left( x\right) = 2... | Yes |
Lemma 7.8.1. At points of inflection, \( {f}^{\prime \prime }\left( x\right) = 0 \) . | Solving the equation resulting from Lemma 7.8.1 gives the \( x \) -coordinate(s) of the point(s) of inflection. To determine the direction of curving or direction of bending in the vicinity of a point of inflection, test \( {f}^{\prime \prime }\left( x\right) \) for values of \( x \), first slightly less and then sligh... | Yes |
Find the differential of the arc of the circle \( {x}^{2} + {y}^{2} = {r}^{2} \) . | Differentiating, \( \frac{dy}{dx} = - \frac{x}{y} \) . To find \( {ds} \) in terms of \( x \) we substitute in (8.6), giving \[ {ds} = {\left\lbrack 1 + \frac{{x}^{2}}{{y}^{2}}\right\rbrack }^{\frac{1}{2}}{dx} = {\left\lbrack \frac{{y}^{2} + {x}^{2}}{{y}^{2}}\right\rbrack }^{\frac{1}{2}}{dx} = {\left\lbrack \frac{{r}^{... | Yes |
Find the differential of the arclength of the cardioid \( \rho = a(l - \cos \theta ) \) in terms of \( \theta \) . | Solution. Differentiating, \( \frac{d\rho }{d\theta } = a\sin \theta \) .\n\nSubstituting in (8.10), gives\n\n\[ \n{ds} = {\left\lbrack {a}^{2}{\left( 1 - \cos \theta \right) }^{2} + {a}^{2}{\sin }^{2}\theta \right\rbrack }^{\frac{1}{2}}{d\theta } = a{\left\lbrack 2 - 2\cos \theta \right\rbrack }^{\frac{1}{2}}{d\theta ... | Yes |
Change the independent variable from \( x \) to \( y \) in the equation\n\n\[ 3{\left( \frac{{d}^{2}y}{d{x}^{2}}\right) }^{2} - \frac{dy}{dx}\frac{{d}^{3}y}{d{x}^{3}} - \frac{{d}^{2}y}{d{x}^{2}}{\left( \frac{dy}{dx}\right) }^{2} = 0. \] | Solution. Substituting from (9.1), (9.3), (9.4),\n\n\[ 3{\left( -\frac{\frac{{d}^{2}x}{d{y}^{2}}}{{\left( \frac{dx}{dy}\right) }^{3}}\right) }^{2} - \left( \frac{1}{\frac{dx}{dy}}\right) \left( {-\frac{\frac{{d}^{3}x}{d{y}^{3}}\frac{dx}{dy} - 3{\left( \frac{{d}^{2}x}{d{y}^{2}}\right) }^{2}}{{\left( \frac{dx}{dy}\right)... | Yes |
Having given the equation\n\n\[ \frac{{d}^{2}y}{d{x}^{2}} = 1 + \frac{2\left( {1 + y}\right) }{1 + {y}^{2}}{\left( \frac{dy}{dx}\right) }^{2} \]\n\nchange the dependent variable from \( y \) to \( z \) by means of the relation\n\n\[ y = \tan z. \] | Solution. From the above,\n\n\[ \frac{dy}{dx} = {\sec }^{2}\left( z\right) \frac{dz}{dx},\frac{{d}^{2}y}{d{x}^{2}} = {\sec }^{2}\left( z\right) \frac{{d}^{2}z}{d{x}^{2}} + 2{\sec }^{2}\left( z\right) \tan \left( z\right) {\left( \frac{dz}{dx}\right) }^{2}, \]\n\nSubstituting,\n\n\[ {\sec }^{2}\left( z\right) \frac{{d}^... | Yes |
Change the independent variable from \( x \) to \( t \) in the differential equation\n\n\[ \n{x}^{2}\frac{{d}^{2}y}{d{x}^{2}} + x\frac{dy}{dx} + y = 0 \n\]\n\nwhere \( x = {e}^{t} \) . | Solution. \( \frac{dx}{dt} = {e}^{t} \), therefore\n\n\[ \n\frac{dt}{dx} = {e}^{-t}.\n\]\n\nAlso \( \frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx} \) ; therefore \( \frac{dy}{dx} = {e}^{-t}\frac{dy}{dt} \) . Also \( \frac{{d}^{2}y}{d{x}^{2}} = {e}^{-t}\frac{d}{dx}\left( \frac{dy}{dt}\right) - \frac{dy}{dt}{e}^{-t}\frac{dt}... | Yes |
Transform the formula for the radius of curvature (11.5), \[ R = \frac{{\left\lbrack 1 + {\left( \frac{dy}{dx}\right) }^{2}\right\rbrack }^{\frac{3}{2}}}{\frac{{d}^{2}y}{d{x}^{2}}} \] into polar coordinates. | Solution. Since in (9.5) and (9.6), \( t \) is any variable on which \( x \) and \( y \) depend, we may in this case let \( t = \theta \), giving \( \frac{dy}{dx} = \frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }} \), and \[ \frac{{d}^{2}y}{d{x}^{2}} = \frac{\frac{dx}{d\theta }\frac{{d}^{2}y}{d{\theta }^{2}} - \frac{dy}... | Yes |
Examine \( {x}^{3} - 9{x}^{2} + {24x} - 7 \) for maximum and minimum values. | Solution. \( f\left( x\right) = {x}^{3} - 9{x}^{2} + {24x} - 7.{f}^{\prime }\left( x\right) = 3{x}^{2} - {18x} + {24} \) . Solving \( 3{x}^{2} - {18x} + {24} = 0 \) gives the critical values \( x = 2 \) and \( x = 4 \) . Thus \( {f}^{\prime }\left( 2\right) = 0 \) , and \( {f}^{\prime }\left( 4\right) = 0 \) . Differen... | Yes |
Example 10.5.2. Examine \( {e}^{x} + 2\cos \left( x\right) + {e}^{-x} \) for maximum and minimum values. | Solution. \( f\left( x\right) = {e}^{x} + 2\cos \left( x\right) + {e}^{-x},{f}^{\prime }\left( x\right) = {e}^{x} - 2\sin x - {e}^{-x} = 0 \), for \( x = 0 \) (and \( x = 0 \) is the only root of the equation \( {e}^{x} - 2\sin x - {e}^{-x} = 0 \) ), \( {f}^{\prime \prime }\left( x\right) = {e}^{x} - 2\cos \left( x\rig... | Yes |
For \( x = 2 \) the function \( \frac{{x}^{2} - 4}{x - 2} \) assumes the form \( \frac{0}{0} \) but | \[ \mathop{\lim }\limits_{{x \rightarrow 2}}\frac{{x}^{2} - 4}{x - 2} = 4 \] | No |
Consider the three functions \( \frac{\sin x}{x},\frac{\sin x}{\left| x\right| } \) and \( \frac{\sin x}{1 - \cos x} \) at the point \( x = 0 \). | \[ \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\sin x}{x} = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\cos x}{1} = 1 \] Thus \( \frac{\sin x}{x} \) has a removable discontinuity at \( x = 0 \). \[ \mathop{\lim }\limits_{{x \rightarrow {0}^{ + }}}\frac{\sin x}{\left| x\right| } = \mathop{\lim }\limits_{{x \right... | Yes |
We use Sage to compute \( \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\cos \left( x\right) - 1}{{x}^{2}} \) . | sage: limit((cos(x)-1)/x^2, x=0)\n\n\( - 1/2 \)\n\nsage: limit((-sin(x))/(2*x), x=0)\n\n\( - 1/2 \)\n\nsage: limit((-cos(x))/(2), x=0)\n\n\( - 1/2 \)\n\nThis verifies\n\n\[ \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\cos \left( x\right) - 1}{{x}^{2}} = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{-\sin \left( x\r... | Yes |
Let \( a \) and \( d \) be nonzero.\n\n\[ \mathop{\lim }\limits_{{x \rightarrow \infty }}\frac{a{x}^{2} + {bx} + c}{d{x}^{2} + {ex} + f} = \mathop{\lim }\limits_{{x \rightarrow \infty }}\frac{{2ax} + b}{{2dx} + e} \] | \[ = \mathop{\lim }\limits_{{x \rightarrow \infty }}\frac{2a}{2d} \]\n\n\[ = \frac{a}{d} \] | Yes |
Consider\n\n\\[ \n\\mathop{\\lim }\\limits_{{x \\rightarrow 0}}\\frac{\\cos x - 1}{x\\sin x} \n\\] | This limit is an indeterminate of the form \\( \\frac{0}{0} \\) . Applying L’Hospital’s rule we see that limit is equal to\n\n\\[ \n\\mathop{\\lim }\\limits_{{x \\rightarrow 0}}\\frac{-\\sin x}{x\\cos x + \\sin x}.\n\\]\n\nThis limit is again an indeterminate of the form \\( \\frac{0}{0} \\) . We apply L’Hospital’s rul... | Yes |
We use Sage to compute \( \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\cos \left( x\right) - 1}{{x}^{2}} \) . | sage: limit((cos(x)-1)/x^2, x=0)\n\n\( - 1/2 \)\n\nsage: limit \( \left( {\left( {-\sin \left( x\right) }\right) /\left( {2 * x}\right), x = 0}\right) \)\n\n\( - 1/2 \)\n\nsage: limit((-cos(x))/(2), x=0)\n\n\( - 1/2 \)\n\nThis verifies\n\n\[ \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\cos \left( x\right) - 1}{{x}^{... | Yes |
Evaluate \( \frac{f\left( x\right) }{g\left( x\right) } = \frac{{x}^{3} - {3x} + 2}{{x}^{3} - {x}^{2} - x - 1} \) when \( x = 1 \) . | \[ {\left. \frac{f\left( 1\right) }{g\left( 1\right) } = \frac{{x}^{3} - {3x} + 2}{{x}^{3} - {x}^{2} + 1}\right\rbrack }_{x = 1} = \frac{1 - 3 + 2}{1 - 1 - 1 + 1} = \frac{0}{0}. \] Therefore, this is an indeterminate form. \[ {\left. \frac{{f}^{\prime }\left( 1\right) }{{g}^{\prime }\left( 1\right) } = \frac{3{x}^{2} -... | Yes |
Evaluate \( \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{{e}^{x} - {e}^{-x} - {2x}}{x - \sin x} \) . | \[ {\left. \frac{f\left( 0\right) }{g(0} = \frac{{e}^{x} - {e}^{-x} - {2x}}{x - \sin x}\right\rbrack }_{x = 0} = \frac{1 - 1 - 0}{0 - 0} = \frac{0}{0}. \] Therefore, this is an indeterminate form. \[ {\left. \frac{{f}^{\prime }\left( 0\right) }{{g}^{\prime }\left( 0\right) } = \frac{{e}^{x} - {e}^{-x} - 2}{1 - \cos x}\... | Yes |
Evaluate \( \frac{\log x}{\csc x} \) for \( x = 0 \) . | \[ \n{\left. \frac{f\left( 0\right) }{g\left( 0\right) } = \frac{\log \left( x\right) }{\csc \left( x\right) }\right\rbrack }_{x = 0} = \frac{-\infty }{\infty }.\n\]\n\nTherefore, this is an indeterminate form.\n\n\[ \n{\left. \frac{{f}^{\prime }\left( 0\right) }{{g}^{\prime }\left( 0\right) } = \frac{\frac{1}{x}}{-\cs... | Yes |
Let \( a \) and \( d \) be nonzero.\n\n\[ \mathop{\lim }\limits_{{x \rightarrow \infty }}\frac{a{x}^{2} + {bx} + c}{d{x}^{2} + {ex} + f} = \mathop{\lim }\limits_{{x \rightarrow \infty }}\frac{{2ax} + b}{{2dx} + e} \] | \[ = \mathop{\lim }\limits_{{x \rightarrow \infty }}\frac{2a}{2d} \]\n\n\[ = \frac{a}{d} \] | Yes |
Evaluate \( \sec \left( {3x}\right) \cos \left( {5x}\right) \) for \( x = \frac{\pi }{2} \). | Solution. \( {\left. \sec 3x\cos 5x\right\rbrack }_{x = \frac{\pi }{2}} = \infty \cdot 0 \) . Therefore, this is an indeterminate form. Substituting \( \frac{1}{\cos {3x}} \) for \( \sec {3x} \), the function becomes \( \frac{\cos {5x}}{\cos {3x}} = \frac{f\left( x\right) }{g\left( x\right) } \). \n\n\[ \n{\left. \frac... | Yes |
\[ \mathop{\lim }\limits_{{x \rightarrow 0}}\left( {\cot x - \frac{1}{x}}\right) = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{x\cos x - \sin x}{x\sin x} \] | \[ = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\cos x - x\sin x - \cos x}{\sin x + x\cos x} \] \[ = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{-x\sin x}{\sin x + x\cos x} \] \[ = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{-x\cos x - \sin x}{\cos x + \cos x - x\sin x} \] \[ = 0 \] | No |
Evaluate \( \sec x - \tan x \) for \( x = \frac{\pi }{2} \). | Solution. \( {\left. \sec x - \tan x\right\rbrack }_{x = \frac{\pi }{2}} = \infty - \infty \) . Therefore, this is an indeterminate form. By Trigonometry,\n\n\( \sec x - \tan x = \frac{1}{\cos x} - \frac{\sin x}{\cos x} = \frac{1 - \sin x}{\cos x} = \frac{f\left( x\right) }{g\left( x\right) } \) .\n\n\( \frac{f\left( \... | Yes |
Evaluate \( {x}^{x} \) when \( x = 0 \) . | This function assumes the indeterminate form \( {0}^{0} \) for \( x = 0 \) . Let \( y = {x}^{x} \) ; then \( \log y = x\log x = 0 \cdot \left( {-\infty }\right) \), when \( x = 0 \) . By \( §{10.11} \) , \n\n\[ \n\log y\frac{\log x}{\frac{1}{x}} = \frac{-\infty }{\infty } \n\] \n\nwhen \( x = 0 \) . By \( §{10.10} \) ,... | No |
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