Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Example 3.12. Let \( f \) be the function given by the following graph. | Solution. Most of the answers can be read immediately from the graph.\n\n(a) For the domain, we project the graph to the \( x \) -axis. The domain consists of all numbers from -5 to 5 without -3, that is \( D = \left\lbrack {-5, - 3)\cup ( - 3,5}\right\rbrack \) .\n\n(b) For the range, we project the graph to the \( y ... | Yes |
a) What was the population size in the years 2004 and 2009?\nb) In what years did the city have the most population?\nc) In what year did the population grow the fastest?\nd) In what year did the population decline the fastest? | Solution. The population size in the year 2004 was approximately 36,000 . In the year 2009, it was approximately 26,000 . The largest population was in the year 2006, where the graph has its maximum. The fastest growth in the population was between the years 2003 and 2004. That is when the graph has the largest slope. ... | Yes |
Graph the function described by the following formula:\n\n\\[ f\\left( x\\right) = \\left\\{ \\begin{matrix} x + 3 & ,\\text{ for } & - 3 \\leq x < - 1 \\\\ {x}^{2} & ,\\text{ for } & - 1 < x < 1 \\\\ 3 & ,\\text{ for } & 2 < x \\leq 3 \\end{matrix}\\right. \\] | Solution. We really have to graph all three functions \\( y = x + 3 \\) , \\( y = {x}^{2} \\), and \\( y = 3 \\), and then restrict them to their respective domain. Graphing the three functions, we obtain the following tables and associated graphs, which we draw in one \\( x - y \\) -plane:\n\n<table><thead><tr><th col... | No |
Graph the equations \( y = \sqrt{7 - x} \) and \( y = {x}^{3} - 2{x}^{2} - 4 \) in the same window. | Solution. Enter the functions as Y1 and Y2 after pressing \( \left( \overline{y = }\right) \) . The graphs of both functions appear together in the graphing window: | No |
Graph the relation \( {\left( x - 3\right) }^{2} + {\left( y - 5\right) }^{2} = {16} \) . | Since the above expression is not solved for \( y \), we cannot simply plug this into the calculator. Instead, we have to solve for \( y \) first.\n\n\[{\left( x - 3\right) }^{2} + {\left( y - 5\right) }^{2} = {16}\; \Rightarrow \;{\left( y - 5\right) }^{2} = {16} - {\left( x - 3\right) }^{2}\]\n\n\[ \Rightarrow \;y - ... | Yes |
Example 4.6. Graph the equation \( y = {x}^{3} - 2{x}^{2} - {4x} + 4 \). | Solution.\n\na) Graphing the function in the standard window gives the following graph:\n\n\n\n, we can move a cursor to the right and left via We can see several function values as displayed }{x}^{4} + 3{x}^{2} - {5x} - 7 = 0\text{, b)}{x}^{3} - 4 = {7x} - {3}^{x}\text{.} \n\] | Solution. a) We need to find all numbers \( x \) so that \( {x}^{4} + 3{x}^{2} - {5x} - 7 = 0 \) . Note, that these are precisely the zeros of the function \( f\left( x\right) = {x}^{4} + 3{x}^{2} - {5x} - 7 \) , since the zeros are the values \( x \) for which \( f\left( x\right) = 0 \) . Graphing the function \( f\le... | No |
Graph the equations \( y = {x}^{2} - {3x} + 2 \) and \( y = {x}^{3} + 2{x}^{2} - 1 \). | Solution. First, enter the two equations for Y1 and Y2 after pressing the y= ) key. Both graphs are displayed in the graphing window.\n\n\n\na) The procedure for finding the intersection of the graphs for Y1 and Y2 is ... | Yes |
Example 5.6. Guess the formula for the function, based on the basic graphs in Section 5.1 and the transformations described above. | Solution. a) This is the square-root function shifted to the left by 2 . Thus, by Observation 5.1, this is the function \( f\left( x\right) = \sqrt{x + 2} \) .\nb) This is the graph of \( y = \frac{1}{x} \) reflected about the \( x \) -axis (or also \( y = \frac{1}{x} \) reflected about the \( y \) -axis). In either ca... | Yes |
Sketch the graph of the function, based on the basic graphs in Section 5.1 and the transformations described above. | Solution. a) This is the parabola \( y = {x}^{2} \) shifted up by 3 . The graph is shown below.\nb) \( y = {\left( x + 2\right) }^{2} \) is the parabola \( y = {x}^{2} \) shifted 2 units to the left.  ![37f2b8bb-bce0-4... | Yes |
a) The graph of \( f\left( x\right) = \left| {{x}^{3} - 5}\right| \) is stretched away from the \( y \) -axis by a factor of 3 . What is the formula for the new function? | a) By Observation 5.4 on page 67 we have to multiply the argument by \( \frac{1}{3} \) . The new function is therefore:\n\n\[ f\left( {\frac{1}{3} \cdot x}\right) = \left| {{\left( \frac{1}{3} \cdot x\right) }^{3} - 5}\right| = \left| {\frac{1}{27} \cdot {x}^{3} - 5}\right| \] | Yes |
Example 5.10. Determine, if the following functions are even, odd, or neither. | Solution. The function \( f\left( x\right) = {x}^{2} \) is even, since \( f\left( {-x}\right) = {\left( -x\right) }^{2} = {x}^{2} \) . Similarly, \( g\left( x\right) = {x}^{3} \) is odd, \( h\left( x\right) = {x}^{4} \) is even, and \( k\left( x\right) = {x}^{5} \) is odd, since\n\n\[ g\left( {-x}\right) = {\left( -x\r... | Yes |
Example 6.1. Let \( f\left( x\right) = {x}^{2} + {5x} \) and \( g\left( x\right) = {7x} - 3 \) . Find the following functions, and state their domain.\n\n\[ \left( {f + g}\right) \left( x\right) ,\left( {f - g}\right) \left( x\right) ,\left( {f \cdot g}\right) \left( x\right) ,\text{ and }\left( \frac{f}{g}\right) \lef... | Solution. The functions are calculated by adding the functions (or subtracting, multiplying, dividing).\n\n\[ \left( {f + g}\right) \left( x\right) = \left( {{x}^{2} + {5x}}\right) + \left( {{7x} - 3}\right) = {x}^{2} + {12x} - 3, \]\n\n\[ \left( {f - g}\right) \left( x\right) = \left( {{x}^{2} + {5x}}\right) - \left( ... | Yes |
Example 6.3. Let \( f\left( x\right) = \sqrt{x + 2} \), and let \( g\left( x\right) = {x}^{2} - {5x} + 4 \) . Find the functions \( \frac{f}{g} \) and \( \frac{g}{f} \) and state their domains. | Solution. First, the domain of \( f \) consists of those numbers \( x \) for which the square root is defined. In other words, we need \( x + 2 \geq 0 \), that is \( x \geq - 2 \), so that the domain of \( f \) is \( {D}_{f} = \lbrack - 2,\infty ) \) . On the other hand, the domain of \( g \) is all real numbers, \( {D... | Yes |
To form the quotient \( \frac{f}{g}\left( x\right) \) where \( f\left( x\right) = {x}^{2} - 1 \) and \( g\left( x\right) = x + 1 \) | we write \( \frac{f}{g}\left( x\right) = \frac{{x}^{2} - 1}{x + 1} = \frac{\left( {x + 1}\right) \left( {x - 1}\right) }{x + 1} = x - 1 \) . One might be tempted to say that the domain is all real numbers. But it is not! The domain is all real numbers except -1 , and the last step of the simplification performed above ... | Yes |
Example 6.6. Let \( f\left( x\right) = 2{x}^{2} + {5x} \) and \( g\left( x\right) = 2 - x \) . Find the following compositions\n\n\[ \n\\text{a)}f\\left( {g\\left( 3\\right) }\\right) ,\\;\\text{b)}g\\left( {f\\left( 3\\right) }\\right) ,\\;\\text{c)}f\\left( {f\\left( 1\\right) }\\right) ,\\;\\text{d)}f\\left( {2 \\cd... | Solution. We evaluate the expressions, one at a time, as follows:\n\n\[ \n\\text{a)}\\;f\\left( {g\\left( 3\\right) }\\right) = f\\left( {2 - 3}\\right) = f\\left( {-1}\\right) = 2 \\cdot {\\left( -1\\right) }^{2} + 5 \\cdot \\left( {-1}\\right) \n\]\n\n\[ \n= 2 - 5 = - 3 \n\]\n\nb) \\;g\\left( {f\\left( 3\\right) }\\r... | Yes |
Find \( \left( {f \circ g}\right) \left( x\right) \) and \( \left( {g \circ f}\right) \left( x\right) \) for the following functions, and state their domains. | a) Composing \( f \circ g \), we obtain\n\n\[ \left( {f \circ g}\right) \left( x\right) = f\left( {g\left( x\right) }\right) = \frac{3}{g\left( x\right) + 2} = \frac{3}{{x}^{2} - {3x} + 2}. \]\n\nThe domain is the set of numbers \( x \) for which the denominator is non-zero.\n\n\[ {x}^{2} - {3x} + 2 = 0\; \Rightarrow \... | Yes |
Example 6.9. Let \( f \) and \( g \) be the functions defined by the following table.\n\n<table><thead><tr><th>\( x \)</th><th>1</th><th>2</th><th>3</th><th>4</th><th>5</th><th>6</th><th>7</th></tr></thead><tr><td>\( f\left( x\right) \)</td><td>6</td><td>3</td><td>1</td><td>4</td><td>0</td><td>7</td><td>6</td></tr><tr>... | Solution. For (a) and (b), we obtain by immediate calculation\n\n<table><thead><tr><th>\( x \)</th><th>1</th><th>2</th><th>3</th><th>4</th><th>5</th><th>6</th><th>7</th></tr></thead><tr><td>\( 2 \cdot f\left( x\right) + 3 \)</td><td>15</td><td>9</td><td>5</td><td>11</td><td>3</td><td>17</td><td>15</td></tr><tr><td>\( f... | No |
Example 6.10. Let \( f \) and \( g \) be the functions defined by the following table.\n\n<table><thead><tr><th>\( x \)</th><th>1</th><th>3</th><th>5</th><th>7</th><th>9</th><th>11</th></tr></thead><tr><td>\( f\left( x\right) \)</td><td>3</td><td>5</td><td>11</td><td>4</td><td>9</td><td>7</td></tr><tr><td>\( g\left( x\... | Solution. The compositions are calculated by repeated evaluation. For example,\n\n\[ \left( {f \circ g}\right) \left( 1\right) = f\left( {g\left( 1\right) }\right) = f\left( 7\right) = 4. \]\n\nThe complete answer is displayed below.\n\n<table><thead><tr><th>\( x \)</th><th>1</th><th>3</th><th>5</th><th>7</th><th>9</th... | Yes |
As was noted above, the function \( f\left( x\right) = {x}^{2} \) is not one-to-one, because, for example, for inputs 2 and -2 , we have the same output | \[ f\left( {-2}\right) = {\left( -2\right) }^{2} = 4,\;f\left( 2\right) = {2}^{2} = 4. \] | Yes |
c) \( f\left( x\right) = - {x}^{3} + 6{x}^{2} - {13x} + {12} \) | Solution. We use the horizontal line test to see which functions are one-toone. For (a) and (b), we see that the functions are not one-to-one since there\nis a horizontal line that intersects with the graph more than once:  \( f\left( x\right) = {2x} - 7 \) | a) First, reverse the role of input and output in \( y = {2x} - 7 \) by exchanging the variables \( x \) and \( y \) . That is, we write \( x = {2y} - 7 \) . We need to solve this for \( y \) :\n\n\[ \overset{\left( \text{ add }7\right) }{ \Rightarrow }\;x + 7 = {2y} \Rightarrow \;y = \frac{x + 7}{2}. \]\n\nTherefore, ... | Yes |
Note, that the function \( y = {x}^{2} \) can be restricted to a one-to-one function by choosing the domain to be all non-negative numbers \( \lbrack 0,\infty ) \) or by choosing the domain to be all non-positive numbers \( ( - \infty ,0\rbrack \) . | Let \( f : \lbrack 0,\infty ) \rightarrow \lbrack 0,\infty ) \) be the function \( f\left( x\right) = {x}^{2} \), so that \( f \) has a domain of all non-negative numbers. Then, the inverse is the function \( {f}^{-1}\left( x\right) = \sqrt{x} \). On the other hand, we can take \( g\left( x\right) = {x}^{2} \) whose do... | Yes |
Example 7.8. Restrict the function to a one-to-one function. Find the inverse function, if possible. | Solution. The graphs of \( f \) and \( g \) are displayed below.\na)\n\na) The graph shows that \( f \) is one-to-one when restricted to all numbers \( x \geq - 3 \), which is the choice we make to find an inverse fu... | Yes |
Are the following functions inverse to each other? | Solution. We calculate the compositions \( f\left( {g\left( x\right) }\right) \) and \( g\left( {f\left( x\right) }\right) \) .\n\n\[ \text{a)}\;f\left( {g\left( x\right) }\right) = f\left( \frac{x - 7}{5}\right) = 5 \cdot \frac{x - 7}{5} + 7 = \left( {x - 7}\right) + 7 = x\text{,}\]\n\n\[ g\left( {f\left( x\right) }\r... | Yes |
Example 7.13. Find the graph of the inverse function of the function given below.\n\nb) \( f\left( x\right) = {\left( x + 1\right) }^{3} \) | Solution. Carefully reflecting the graphs given in part (a) and (b) gives the following solution. The function \( f\left( x\right) = {\left( x + 1\right) }^{3} \) in part (b) can be graphed with a calculator first. | No |
Divide the following fractions via long division:\n\n\[ \n\\text{a)}\\frac{3571}{11}\\;\\text{b)}\\frac{{x}^{3} + 5{x}^{2} + {4x} + 2}{x + 3} \n\] | Solution. a) Recall the procedure for long division of natural numbers: \n\nThe steps above are performed as follows. First, we find the largest multiple of 11 less or equal to 35 . The answer 3 is written as the fir... | No |
Divide the following fractions via long division. | Solution. For part (a), we calculate:\n\n\n\nTherefore, \( {x}^{2} + {4x} + 5 = \\left( {x + 8}\\right) \\cdot \\left( {x - 4}\\right) + {37} \) . | No |
Find the remainder of dividing \( f\left( x\right) = {x}^{2} + {3x} + 2 \) by\n\n\[ \text{a)}x - 3\text{, b)}x + 4\text{, c)}x + 1\text{, d)}x - \frac{1}{2}\text{.} \] | Solution. By Observation 8.10 we know that the remainder \( r \) of the division by \( x - c \) is \( f\left( c\right) \) . Thus, the remainder for part (a), when dividing by \( x - 3 \) is\n\n\[ r = f\left( 3\right) = {3}^{2} + 3 \cdot 3 + 2 = 9 + 9 + 2 = {20}. \]\n\nFor (b), note that \( g\left( x\right) = x + 4 = x ... | Yes |
Determine whether \( g\left( x\right) \) is a factor of \( f\left( x\right) \). \[ \text{a)}f\left( x\right) = {x}^{3} + 2{x}^{2} + {5x} + 1,\;g\left( x\right) = x - 2\text{,} \] | Solution. a) We need to determine whether 2 is a root of \( f\left( x\right) = {x}^{3} + 2{x}^{2} + {5x} + 1 \), that is, whether \( f\left( 2\right) \) is zero. \[ f\left( 2\right) = {2}^{3} + 2 \cdot {2}^{2} + 5 \cdot 2 + 1 = 8 + 8 + {10} + 1 = {27}. \] Since \( f\left( 2\right) = {27} \neq 0 \), we see that \( g\lef... | Yes |
a) Show that -2 is a root of \( f\left( x\right) = {x}^{5} - 3 \cdot {x}^{3} + 5{x}^{2} - {12} \), and use this to factor \( f \) . | a) First, we calculate that -2 is a root.\n\n\[ f\left( {-2}\right) = {\left( -2\right) }^{5} - 3 \cdot {\left( -2\right) }^{3} + 5 \cdot {\left( -2\right) }^{2} + {12} = - {32} + {24} + {20} - {12} = 0. \]\n\nSo we can divide \( f\left( x\right) \) by \( g\left( x\right) = x - \left( {-2}\right) = x + 2 \) :\n\n\[ \be... | Yes |
Our first example is the long division of \( \frac{5{x}^{3} + 7{x}^{2} + x + 4}{x + 2} \) . | Here, the first term \( 5{x}^{2} \) of the quotient is just copied from the first term of the dividend. We record this together with the coefficients of the dividend \( 5{x}^{3} + 7{x}^{2} + x + 4 \) and of the divisor \( x + 2 = x - \left( {-2}\right) \) as follows:\n\n\[ - 2\left| {\;\begin{array}{llll} 5 & 7 & 1 & 4... | Yes |
Example 8.15. Find the following quotients via synthetic division.\n\n\\[ \n\\text{a)}\\frac{4{x}^{3} - 7{x}^{2} + {4x} - 8}{x - 4}\\text{, b)}\\frac{{x}^{4} - {x}^{2} + 5}{x + 3}\\text{.} \n\\] | Solution. a) We need to perform the synthetic division. \n\nTherefore we have\n\n\\[ \n\\frac{4{x}^{3} - 7{x}^{2} + {4x} - 8}{x - 4} = 4{x}^{2} + {9x} + {40} + \\frac{152}{x - 4}.\n\\] | Yes |
Example 9.8. Which of the following graphs could be the graphs of a polynomial? If the graph could indeed be a graph of a polynomial then determine a possible degree of the polynomial. | ## Solution.\na) Yes, this could be a polynomial. The degree could be, for example, 4.\nb) No, since the graph has a pole.\nc) Yes, this could be a polynomial. A possible degree would be degree 3 .\nd) No, since the graph has a corner.\ne) No, since \( f\left( x\right) \) does not approach \( \infty \) or \( - \infty \... | Yes |
Identify the graphs of the polynomials in (a), (b) and (c) with the functions (i), (ii), and (iii). | Solution. The only odd degree function is (i) and it must therefore correspond to graph (c). For (ii), since the leading coefficient is negative, we know that the function opens downwards, so that it corresponds to graph (a). Finally (iii) opens upwards, since the leading coefficient is positive, so that its graph is \... | Yes |
Example 9.10. Graph of the given function with the TI-84. Include all extrema and intercepts of the graph in your viewing window.\na) \( f\left( x\right) = - {x}^{3} + {26}{x}^{2} - {129x} + {175} \) | Solution. a) The graph in the standard window looks as follows:\n\n\n\nHowever, since the function is of degree 3 , this cannot be the full graph, as \( f\left( x\right) \) has to approach \( - \infty \) when \( x \)... | Yes |
Example 9.11. Find the roots of the polynomial from its graph. | Solution. a) We start by graphing the polynomial \( f\left( x\right) = {x}^{3} - 7{x}^{2} + {14x} - 8 \) .\n\n\n\nThe graph suggests that the roots are at \( x = 1, x = 2 \), and \( x = 4 \) . This may easily be chec... | Yes |
The solutions of the equation \( a{x}^{2} + {bx} + c = 0 \) for some real numbers \( a, b \), and \( c \) are given by | \[ {x}_{1} = \frac{-b + \sqrt{{b}^{2} - {4ac}}}{2a}\text{, and}{x}_{2} = \frac{-b - \sqrt{{b}^{2} - {4ac}}}{2a}\text{.} \] We may combine the two solutions \( {x}_{1} \) and \( {x}_{2} \) and simply write this as: \[ {x}_{1/2} = \frac{-b \pm \sqrt{{b}^{2} - {4ac}}}{2a} \] | Yes |
Consider the equation \( {10}{x}^{3} - 6{x}^{2} + {5x} - 3 = 0 \) . Let \( x \) be a rational solution of this equation, that is \( x = \frac{p}{q} \) is a rational number such that\n\n\[ {10} \cdot {\left( \frac{p}{q}\right) }^{3} - 6 \cdot {\left( \frac{p}{q}\right) }^{2} + 5 \cdot \frac{p}{q} - 3 = 0. \] | We assume that \( x = \frac{p}{q} \) is completely reduced, that is, \( p \) and \( q \) have no common factors that can be used to cancel the numerator and denominator of the fraction \( \frac{p}{q} \) . Now, simplifying the above equation, and combining terms, we obtain:\n\n\[ {10} \cdot \frac{{p}^{3}}{{q}^{3}} - 6 \... | Yes |
a) Find all rational roots of \( f\left( x\right) = 7{x}^{3} + {x}^{2} + {7x} + 1 \) . | a) If \( x = \frac{p}{a} \) is a rational root, then \( p \) is a factor of 1, that is \( p = \pm 1 \) , and \( q \) is a factor of \( 7 \), that is \( q = \pm 1, \pm 7 \) . The candidates for rational roots are therefore \( x = \pm \frac{1}{1}, \pm \frac{1}{7} \) . To see which of these candidates are indeed roots of ... | Yes |
Find roots of the given polynomial and use this information to factor the polynomial completely. | a) In order to find a root, we use the graph to make a guess for one of the roots.\n\n\n\nThe graph suggests that the roots may be at \( x = - 2 \) and \( x = 3 \) . This is also supported by looking at the table for... | Yes |
Find the roots of the polynomial and sketch its graph including all roots. | a) To find a root, we first graph the function \( f\left( x\right) = {x}^{3} + 2{x}^{2} - {14x} - 3 \) with the calculator. The graph and the table suggest that we have a root at \( x = 3 \). Therefore we divide \( f\left( x\right) \) by \( \left( {x - 3}\right) \). We obtain: This shows that \( f\left( x\right) = \lef... | Yes |
Find a polynomial \( f \) with the following properties.\na) \( f \) has degree 3, the roots of \( f \) are precisely \( 4,5,6 \), and the leading coefficient of \( f \) is 7 | In general a polynomial \( f \) of degree 3 is of the form \( f\left( x\right) = \) \( m \cdot \left( {x - {c}_{1}}\right) \cdot \left( {x - {c}_{2}}\right) \cdot \left( {x - {c}_{3}}\right) \) . Identifying the roots and the leading coefficient, we obtain the polynomial\n\n\[ f\left( x\right) = 7 \cdot \left( {x - 4}\... | Yes |
a) Our first graph is \( f\left( x\right) = \frac{1}{x - 3} \) . | Here, the domain is all numbers where the denominator is not zero, that is \( D = \mathbb{R} - \{ 3\} \) . There is a vertical asymptote, \( x = 3 \) . Furthermore, the graph approaches 0 as \( x \) approaches \( \pm \infty \) . Therefore, \( f \) has a horizontal asymptote, \( y = 0 \) . Indeed, whenever the denominat... | Yes |
Example 11.5. Find the domain, all horizontal asymptotes, vertical asymptotes, removable singularities, and \( x \) - and \( y \) -intercepts. Use this information together with the graph of the calculator to sketch the graph of \( f \) . | Solution. a) We combine our knowledge of rational functions and its algebra with the particular graph of the function. The calculator gives the following graph.\n\n\n\nTo find the domain of \( f \) we only need to ex... | Yes |
Graph the function \( p\left( x\right) = \frac{-3{x}^{2}{\left( x - 2\right) }^{3}\left( {x + 2}\right) }{\left( {x - 1}\right) {\left( x + 1\right) }^{2}{\left( x - 3\right) }^{3}} \) . | Solution. We can see that \( p \) has zeros at \( x = 0,2 \), and -2 and vertical asymptotes \( x = 1 \) , \( x = - 1 \) and \( x = 3 \) . Also note that for large \( \left| x\right| \) , \( p\left( x\right) \approx - 3 \) . So there is a horizontal asymptote \( y = - 3 \) . We indicate each of these facts\non the grap... | Yes |
Example 11.7. Sketch the graph of\n\n\[ r\left( x\right) = \frac{2{x}^{2}{\left( x - 1\right) }^{3}\left( {x + 2}\right) }{{\left( x + 1\right) }^{4}{\left( x - 2\right) }^{3}}. \] | Solution. Here we see that there are \( x \) -intercepts at \( \left( {0,0}\right) ,\left( {0,1}\right) \), and \( \left( {0, - 2}\right) \). There are two vertical asymptotes: \( x = - 1 \) and \( x = 2 \). In addition, there is a horizontal asymptote at \( y = 0 \). (Why?) Putting this information on the graph gives\... | Yes |
Solve for \( x \). a) \( -3x + 7 > 19 \), b) \( 2x + 5 \geq 4x - 11 \) | Solution. The first three calculations are straightforward.\n\n\[ \text{a)}\; - {3x} + 7 > {19}\;\overset{\left( -7\right) }{ \Rightarrow }\; - {3x} > {12}\overset{\left( \div \left( -3\right) \right) }{ \Rightarrow }\;x < - 4 \]\n\n\[ \text{b)}\;{2x} + 5 \geq {4x} - {11}\;\overset{\left( -4x - 5\right) }{ \Rightarrow ... | Yes |
Example 12.2. Solve for \( x \). a) \( {x}^{2} - {3x} - 4 \geq 0 \) | Solution. a) We can find the roots of the polynomial on the left by factoring.\n\n\[ {x}^{2} - {3x} - 4 = 0 \Rightarrow \;\left( {x - 4}\right) \left( {x + 1}\right) = 0 \Rightarrow \;x = 4\text{ or }x = - 1 \]\n\nTo see where \( f\left( x\right) = {x}^{2} - {3x} - 4 \) is \( \geq 0 \), we graph it with the calculator.... | Yes |
Find the domain of the given functions.\n\n\[ \text{a)}f\left( x\right) = \sqrt{{x}^{2} - 4}\;\text{b)}g\left( x\right) = \sqrt{{x}^{3} - 5{x}^{2} + {6x}} \] | Solution. a) The domain of \( f\left( x\right) = \sqrt{{x}^{2} - 4} \) is given by all \( x \) for which the square root is non-negative. In other words the domain is given by numbers \( x \) with \( {x}^{2} - 4 \geq 0 \) . Graphing the function \( y = {x}^{2} - 4 = \left( {x + 2}\right) \left( {x - 2}\right) \), we se... | No |
a) \( \frac{{x}^{2} - {5x} + 6}{{x}^{2} - {5x}} \geq 0 \) | Here is the graph of \( \frac{{x}^{2} - {5x} + 6}{{x}^{2} - {5x}} \) in the standard window.\n\n\n\nFactoring numerator and denominator, we can determine vertical asymptotes, holes, and \( x \) -intercepts.\n\n\[ \fr... | Yes |
Graph the functions \[ f\left( x\right) = {2}^{x},\;g\left( x\right) = {3}^{x},\;h\left( x\right) = {10}^{x},\;k\left( x\right) = {\left( \frac{1}{2}\right) }^{x},\;l\left( x\right) = {\left( \frac{1}{10}\right) }^{x}. \] | First, we will graph the function \( f\left( x\right) = {2}^{x} \) by calculating the function values in a table and then plotting the points in the \( x - y \) -plane. We can calculate the values by hand, or simply use the table function of the calculator to find the function values.\n\n\[ f\left( 0\right) = {2}^{0} =... | Yes |
Example 13.5. Graph the functions.\n\n\\[ \n\\text{a)}y = {e}^{x}\\text{, b)}y = {e}^{-x}\\text{, c)}y = {e}^{-{x}^{2}}\\text{, d)}y = \\frac{{e}^{x} + {e}^{-x}}{2} \n\\] | Solution. Using the calculator, we obtain the desired graphs. The exponential function \\( y = {e}^{x} \\) may be entered via 2 nd ln\n\n\n\nNote that the minus sign is entered in the last expression (and also in the... | Yes |
Example 13.6. Graph the functions.\n\n\[ \n\\text{a)}y = {2}^{x}\\text{, b)}y = 3 \\cdot {2}^{x}\\text{, c)}y = \\left( {-3}\\right) \\cdot {2}^{x}\n\]\n\n\[ \n\\text{d)}y = {0.2} \\cdot {2}^{x}\\text{, e)}y = \\left( {-{0.2}}\\right) \\cdot {2}^{x}\n\] | Solution. We graph the functions in one viewing window.\n\n\n\n\n\nHere are the graphs of functions \( f\\left( x\\right) = c \\cdot {2}^{x} \) for various choices of \( c \) .\n\n![37f2b8bb-bce0-4882-a084-b5aa2f5782... | Yes |
Graph the functions. \[ \text{a)}y = {3}^{x} - 5\text{, b)}y = {e}^{x + 4}\text{, c)}y = \frac{1}{4} \cdot {e}^{x - 3} + 2 \] | Solution. The graphs are displayed below. The first graph \( y = {3}^{x} - 5 \) is the graph of \( y = {3}^{x} \) shifted down by 5. | No |
Rewrite the equation as a logarithmic equation. | \[ \text{a)}{3}^{4} = {81}\text{, b)}{10}^{3} = {1000}\text{,} \] \[ \text{c)}{e}^{x} = {17}\text{, d)}{2}^{7 \cdot a} = {53}\text{.} \] Solution. We can immediately apply equation (13.2). For part (a), we have \( b = 3, y = 4 \), and \( x = {81} \) . Therefore we have: \[ {3}^{4} = {81}\; \Leftrightarrow \;{\log }_{3}... | Yes |
Evaluate the expression by rewriting it as an exponential expression.\na) \( {\log }_{2}\left( {16}\right) \) | a) If we set \( y = {\log }_{2}\left( {16}\right) \), then this is equivalent to \( {2}^{y} = {16} \) . Since, clearly, \( {2}^{4} = {16} \), we see that \( y = 4 \) . Therefore, we have \( {\log }_{2}\left( {16}\right) = 4 \) . | Yes |
Example 13.12. Calculate: a) \( {\log }_{3}\left( {13}\right) \) | Solution. We calculate \( {\log }_{3}\left( {13}\right) \) by using the first formula in (13.4).\n\n\[ \n{\log }_{3}\left( {13}\right) = \frac{\log \left( {13}\right) }{\log \left( 3\right) } \approx {2.335} \n\] \n\nAlternatively, we can also calculate this with the second formula in (13.4).\n\n\[ \n{\log }_{3}\left( ... | Yes |
a) Graph the functions \( f\left( x\right) = \ln \left( x\right), g\left( x\right) = \log \left( x\right), h\left( x\right) = {\log }_{2}\left( x\right) \), and \( k\left( x\right) = {\log }_{0.5}\left( x\right) \) . What are the domains of \( f, g, h \), and \( k \) ? How do these functions differ? | Solution. a) We know from the definition that the domain of \( f, g \), and \( h \) is all real positive numbers, \( {D}_{f} = {D}_{g} = {D}_{h} = {D}_{k} = \{ x \mid x > 0\} \) . The functions \( f \) and \( g \) can immediately be entered into the calculator. The standard window gives the following graphs.\n\n![37f2b... | Yes |
Proposition 14.1. The logarithm behaves well with respect to products, quotients, and exponentiation. Indeed, for all positive real numbers \( 0 < b \neq 1 \) , \( x > 0, y > 0 \), and real numbers \( n \), we have:\n\n\[{\log }_{b}\left( {x \cdot y}\right) = {\log }_{b}\left( x\right) + {\log }_{b}\left( y\right)\]\n\... | Proof. We start with the first formula \( {\log }_{b}\left( {x \cdot y}\right) = {\log }_{b}\left( x\right) + {\log }_{b}\left( y\right) \) . If we call \( u = {\log }_{b}\left( x\right) \) and \( v = {\log }_{b}\left( y\right) \), then the equivalent exponential formulas are \( {b}^{u} = x \) and \( {b}^{v} = y \) . W... | Yes |
Combine the terms using the properties of logarithms so as to write as one logarithm. | a) \( \frac{1}{2}\ln \left( x\right) + \ln \left( y\right) = \ln \left( {x}^{\frac{1}{2}}\right) + \ln \left( y\right) = \ln \left( {{x}^{\frac{1}{2}}y}\right) = \ln \left( {\sqrt{x} \cdot y}\right) \) | Yes |
Example 14.3. Write the expressions in terms of elementary logarithms \( u = \) \( {\log }_{b}\left( x\right), v = {\log }_{b}\left( y\right) \), and, in part (c), also \( w = {\log }_{b}\left( z\right) \) . Assume that \( x, y, z > 0 \) .\n\n\[ \n\text{a)}\ln \left( {\sqrt{{x}^{5}} \cdot {y}^{2}}\right) ,\;\text{b)}\l... | Solution. In a first step, we rewrite the expression with fractional exponents, and then apply the rules from proposition 14.1\n\n\[ \n\text{a)}\ln \left( {\sqrt{{x}^{5}} \cdot {y}^{2}}\right) = \ln \left( {{x}^{\frac{5}{2}} \cdot {y}^{2}}\right) = \ln \left( {x}^{\frac{5}{2}}\right) + \ln \left( {y}^{2}\right) \n\]\n\... | Yes |
Example 14.5. Solve for \( x \). a) \( {2}^{x + 7} = {32} \) | \[ \text{a)}\;{2}^{x + 7} = {32}\; \Rightarrow \;{2}^{x + 7} = {2}^{5} \Rightarrow \;x + 7 = 5 \Rightarrow \;x = - 2 \] | Yes |
Example 14.6. Solve for \( x \). a) \( {3}^{x + 5} = 8 \) b) \( {13}^{{2x} - 4} = 6 \) c) \( {5}^{x - 7} = {2}^{x} \) d) \( {5.1}^{x} = {2.7}^{{2x} + 6} \) e) \( {17}^{x - 2} = {3}^{x + 4} \) f) \( {7}^{{2x} + 3} = {11}^{{3x} - 6} \) | Solution. We solve these equations by applying a logarithm (both log or ln will work for solving the equation), and then we use the identity \( \log \left( {a}^{x}\right) = \) \( x \cdot \log \left( a\right) \). a) \( {3}^{x + 5} = 8 \Rightarrow \ln {3}^{x + 5} = \ln 8 \Rightarrow \left( {x + 5}\right) \cdot \ln 3 = \l... | No |
Let \( f\left( x\right) = c \cdot {b}^{x} \). Determine the constant \( c \) and base \( b \) under the given conditions.\na) \( f\left( 0\right) = 5,\;f\left( 1\right) = {20} \) | Solution. a) Applying \( f\left( 0\right) = 5 \) to \( f\left( x\right) = c \cdot {b}^{x} \), we get\n\n\[ 5 = f\left( 0\right) = c \cdot {b}^{0} = c \cdot 1 = c \]\n\nIndeed, in general, we always have \( f\left( 0\right) = c \) for any exponential function. The base \( b \) is then determined by substituting the seco... | Yes |
a) What is the mass of the bacteria sample after 4 hours? | a) The formula for the mass \( y \) in grams after \( t \) hours is \( y\left( t\right) = 2 \cdot {1.02}^{t} \) . Therefore, after 4 hours, the mass in grams is\n\n\[ y\left( 4\right) = 2 \cdot {1.02}^{4} \approx {2.16}. \]\n\n | Yes |
a) Assuming an exponential growth for the population size, find the formula for the population depending on the year \( t \) . | a) The growth is assumed to be exponential, so that \( y\left( t\right) = c \cdot {b}^{t} \) describes the population size depending on the year \( t \), where we set \( t = 0 \) corresponding to the year 2000 . Then the example describes \( y\left( 0\right) = c \) as \( c = {12.7} \), which we assume in units of milli... | Yes |
The number of PCs that are sold in the U.S. in the year 2011 is approximately 350 million with a rate of growth of \( {3.6}\% \) per year. Assuming the rate stays constant over the next years, how many PCs will be sold in the year 2015? | Solution. Since the rate of growth is \( r = {3.6}\% = {0.036} \), we obtain a base of \( b = 1 + r = {1.036} \), giving the number of PCs to be modeled by \( c{\left( {1.036}\right) }^{t} \) . If we set \( t = 0 \) for the year 2011, we find that \( c = {350} \), so the number of sales is given by \( y\left( t\right) ... | Yes |
The size of an ant colony is decreasing at a rate of \( 1\% \) per month. How long will it take until the colony has reached \( {80}\% \) of its original size? | Solution. Since \( r = - 1\% = - {0.01} \), we obtain the base \( b = 1 + r = 1 - {0.01} = \) 0.99 . We have a colony size of \( y\left( t\right) = c \cdot {0.99}^{t} \) after \( t \) months, where \( c \) is the original size. We need to find \( t \) so that the size is at \( {80}\% \) of its original size \( c \), th... | Yes |
The population size of a country is increasing at a rate of \( 4\% \) per year. How long does it take until the country has doubled its population size? | The rate of change is \( r = 4\% = {0.04} \), so that population size is an exponential function with base \( b = 1 + {0.04} = {1.04} \) . Therefore, \( f\left( x\right) = c \cdot {1.04}^{x} \) denotes the population size, with \( c \) being the initial population in the year corresponding to \( x = 0 \) . In order for... | Yes |
Consider the function \( f\left( x\right) = {200} \cdot {\left( \frac{1}{2}\right) }^{\frac{x}{7}} \). We calculate the function values \( f\left( x\right) \), for \( x = 0,7,{14},{21} \), and 28. | \[ f\left( 0\right) = {200} \cdot {\left( \frac{1}{2}\right) }^{\frac{0}{7}} = {200} \cdot 1 = {200} \] \[ f\left( 7\right) = {200} \cdot {\left( \frac{1}{2}\right) }^{\frac{7}{7}} = {200} \cdot \frac{1}{2} = {100} \] \[ f\left( {14}\right) = {200} \cdot {\left( \frac{1}{2}\right) }^{\frac{14}{7}} = {200} \cdot \frac{1... | Yes |
a) Chromium-51 has a half-life of 27.7 days1. How much of 3 grams of chromium-51 will remain after 90 days? | Solution. a) We use the above formula \( y = c \cdot {\left( \frac{1}{2}\right) }^{\frac{t}{h}} \), where \( c = 3 \) grams is the initial amount of chromium-51, \( h = {27.7} \) days is the half-life of chromium-51, and \( t = {90} \) days is time that the isotope decayed. Substituting these numbers into the formula f... | Yes |
We invest an initial amount of \( P = \$ {500} \) for 1 year at a rate of \( r = 6\% \) . The initial amount \( P \) is also called the principal. After 1 year, we receive the principal \( P \) together with the interest \( r \) . \( P \) generated from the principal. The final amount \( A \) after 1 year is therefore | \[ A = \$ {500} + 6\% \cdot \$ {500} = \$ {500} \cdot \left( {1 + {0.06}}\right) = \$ {530}. \] | Yes |
Determine the final amount received on an investment under the given conditions.\na) \$700, compounded monthly, at 4%, for 3 years | Solution. We can immediately use the formula by substituting the given values. For part (a), we have \( P = {700}, n = {12}, r = 4\% = {0.04} \), and \( t = 3 \) . Therefore, we calculate\n\n\[ A = {700} \cdot {\left( 1 + \frac{0.04}{12}\right) }^{{12} \cdot 3} = {700} \cdot {\left( 1 + \frac{0.04}{12}\right) }^{36} \a... | Yes |
a) Find the amount \( P \) that needs to be invested at \( {4.275}\% \) compounded annually for 5 years to give a final amount of \( \$ {3000} \) . (This amount \( P \) is also called the present value of the future amount of \( \$ {3000} \) in 5 years.) | a) We have the following data: \( r = {4.275}\% = {0.04275}, n = 1 \) , \( t = 5 \), and \( A = {3000} \) . We want to find the present value \( P \) . Substituting the given numbers into the appropriate formula, we can solve this for \( P \) .\n\n\[ \n{3000} = P \cdot {\left( 1 + \frac{0.04275}{1}\right) }^{1.5} \Righ... | Yes |
Find \( \sin \left( x\right) ,\cos \left( x\right) \), and \( \tan \left( x\right) \) for the angles\n\n\[ x = {30}^{ \circ },\;x = {45}^{ \circ },\;x = {60}^{ \circ },\;x = {90}^{ \circ },\;x = {0}^{ \circ }. \] | Solution. We draw the angle \( x = {30}^{ \circ } \) in the plane, and use the special triangle to find a point on the terminal side of the angle. From this, we then read off the trigonometric functions. Here are the function values for \( {30}^{ \circ },{45}^{ \circ } \) , and \( {60}^{ \circ } \) .\n\n\[ \sin \left( ... | Yes |
Find \( \sin \left( x\right) ,\cos \left( x\right) \), and \( \tan \left( x\right) \) for the following angles.\n\n\[ \text{a)}x = {240}^{ \circ }\text{，b)}x = {495}^{ \circ }\text{，c)}x = \frac{11\pi }{6}\text{，d)}x = \frac{-{9\pi }}{4} \] | Solution. a) We graph the angle \( x = {240}^{ \circ } \), and identify a special \( {30}^{ \circ } - {60}^{ \circ } - \) \( {90}^{ \circ } \) triangle using the terminal side of \( x \) .\n\n\n\nWe identify a point ... | No |
We graph the functions \( y = \sin \left( x\right), y = \cos \left( x\right) \), and \( y = \tan \left( x\right) \) . | One way to proceed is to calculate and collect various function values in a table and then graph them. However, this is quite elaborate, as the following table shows.\n\n<table><tr><td>\( x \)</td><td>0</td><td>\( \frac{\pi }{6} \)</td><td>\( \frac{\pi }{4} \)</td><td>\( \frac{\pi }{3} \)</td><td>\( \frac{\pi }{2} \)</... | Yes |
Graph the functions:\n\n\[ f\left( x\right) = \sin \left( x\right) + 3, g\left( x\right) = 4 \cdot \sin \left( x\right), h\left( x\right) = \sin \left( {x + 2}\right), i\left( x\right) = \sin \left( {3x}\right) ,\nj\left( x\right) = 2 \cdot \cos \left( x\right) + 3, k\left( x\right) = \cos \left( {{2x} - \pi }\right), ... | Solution. The functions \( f, g, h \), and \( i \) have graphs that are variations of the basic \( y = \sin \left( x\right) \) graph. The graph of \( f\left( x\right) = \sin \left( x\right) + 3 \), shifts the graph of \( y = \sin \left( x\right) \) up by 3, whereas the graph of \( g\left( x\right) = 4 \cdot \sin \left(... | No |
Example 17.10. Find the amplitude, period, and phase-shift, and sketch the graph over one full period. | Solution. a) The amplitude is \( \\left| 3\\right| = 3 \), and since \( f\\left( x\\right) = 3 \\cdot \\sin \\left( {2 \\cdot x + 0}\\right) \), it is \( b = 2 \) and \( c = 0 \), so that the period is \( \\left| \\frac{2\\pi }{2}\\right| = \\pi \) and the phase-shift is \( \\frac{-0}{2} = 0 \) . This is also supported... | Yes |
Proposition 18.1. For any angles \( \alpha \) and \( \beta \), we have\n\n\[ \sin \left( {\alpha + \beta }\right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \] | Proof. We start with the proof of the formulas for \( \sin \left( {\alpha + \beta }\right) \) and \( \cos \left( {\alpha + \beta }\right) \) when \( \alpha \) and \( \beta \) are angles between 0 and \( \frac{\pi }{2} = {90}^{ \circ } \) . We prove the addition formulas (for \( \alpha ,\beta \in \left( {0,\frac{\pi }{2... | No |
Example 18.2. Find the exact values of the trigonometric functions.\n\n\\[ \n\\text{a)}\\cos \\left( \\frac{\\pi }{12}\\right) \\;\\text{b)}\\tan \\left( \\frac{5\\pi }{12}\\right) \\;\\text{c)}\\cos \\left( \\frac{11\\pi }{12}\\right) \n\\] | Solution. a) The key is to realize the angle \\( \\frac{\\pi }{12} \\) as a sum or difference of angles with known trigonometric function values. Note, that \\( \\frac{\\pi }{3} - \\frac{\\pi }{4} = \\frac{{4\\pi } - {3\\pi }}{12} = \\frac{\\pi }{12} \\), so that\n\n\\[ \n\\cos \\left( \\frac{\\pi }{12}\\right) = \\cos... | Yes |
We rewrite \( \cos \left( {x + \frac{\pi }{2}}\right) \) by using the subtraction formula. | \n\[
\cos \left( {x + \frac{\pi }{2}}\right) = \cos x \cdot \cos \frac{\pi }{2} - \sin x \cdot \sin \frac{\pi }{2} = \cos x \cdot 0 + \sin x \cdot 1 = \sin \left( x\right)
\] | Yes |
Find the trigonometric functions using the half-angle formulas. | \[ \text{a)}\sin \left( \frac{\pi }{8}\right) \;\text{b)}\cos \left( \frac{9\pi }{8}\right) \;\text{c)}\tan \left( \frac{\pi }{24}\right) \] Solution. a) Since \( \frac{\pi }{8} = \frac{\frac{\pi }{4}}{2} \), we use the half-angle formula with \( \alpha = \frac{\pi }{4} \). \[ \sin \left( \frac{\pi }{8}\right) = \sin \... | Yes |
Example 18.7. Find the trigonometric functions of \( {2\alpha } \) when \( \alpha \) has the properties below.\n\na) \( \sin \left( \alpha \right) = \frac{3}{5} \), and \( \alpha \) is in quadrant II\n\nb) \( \tan \left( \alpha \right) = \frac{12}{5} \), and \( \alpha \) is in quadrant III | Solution. a) From \( {\sin }^{2}\left( \alpha \right) + {\cos }^{2}\left( \alpha \right) = 1 \), we find that \( {\cos }^{2}\left( \alpha \right) = 1 - {\sin }^{2}\left( \alpha \right) \),\n\nand since \( \alpha \) is in the second quadrant, \( \cos \left( \alpha \right) \) is negative, so that\n\n\[ \cos \left( \alpha... | Yes |
Recall the exact values of the tangent function from section 17.1 | <table><tr><td>\\( x \\)</td><td>\\( 0 = {0}^{ \\circ } \\)</td><td>\\( \\frac{\\pi }{6} = {30}^{ \\circ } \\)</td><td>\\( \\frac{\\pi }{4} = {45}^{ \\circ } \\)</td><td>\\( \\frac{\\pi }{3} = {60}^{ \\circ } \\)</td><td>\\( \\frac{\\pi }{2} = {90}^{ \\circ } \\)</td></tr><tr><td>\\( \\tan \\left( x\\right) \\)</td><td... | Yes |
Example 19.7. We first recall the known values of the sine. | <table><tr><td>\\( x \\)</td><td>\\( 0 = {0}^{ \\circ } \\)</td><td>\\( \\frac{\\pi }{6} = {30}^{ \\circ } \\)</td><td>\\( \\frac{\\pi }{4} = {45}^{ \\circ } \\)</td><td>\\( \\frac{\\pi }{3} = {60}^{ \\circ } \\)</td><td>\\( \\frac{\\pi }{2} = {90}^{ \\circ } \\)</td></tr><tr><td>\\( \\sin \\left( x\\right) \\)</td><td... | Yes |
Example 19.10. We first recall the known values of the cosine. | <table><tr><td>\\( x \\)</td><td>\\( 0 = {0}^{ \\circ } \\)</td><td>\\( \\frac{\\pi }{6} = {30}^{ \\circ } \\)</td><td>\\( \\frac{\\pi }{4} = {45}^{ \\circ } \\)</td><td>\\( \\frac{\\pi }{3} = {60}^{ \\circ } \\)</td><td>\\( \\frac{\\pi }{2} = {90}^{ \\circ } \\)</td></tr><tr><td>\\( \\cos \\left( x\\right) \\)</td><td... | Yes |
Solve for \( x : \;\tan \left( x\right) = \sqrt{3} \) | Solution. There is an obvious solution given by \( x = {\tan }^{-1}\left( \sqrt{3}\right) = \frac{\pi }{3} \), as we studied in the last section. However, we can look for all solutions of \( \tan \left( x\right) = \sqrt{3} \) by studying the graph of the tangent function, that is, by finding all points where the graph ... | Yes |
Solve for \( x \) :\n\n\[ \text{a)}\tan \left( x\right) = 1\text{, b)}\tan \left( x\right) = - 1\text{, c)}\tan \left( x\right) = {5.1}\text{, d)}\tan \left( x\right) = - {3.7} \] | Solution. a) First, we find \( {\tan }^{-1}\left( 1\right) = \frac{\pi }{4} \) . The general solution is thus:\n\n\[ x = \frac{\pi }{4} + n \cdot \pi \;\text{ where }n = 0, \pm 1, \pm 2, \pm 3,\ldots \]\n\n\nb) First, we need to find \( {\tan }^{-1}\left( {-1}\right) \) . Recall from equation (19.1) that \( {\tan }^{-1... | Yes |
Solve for \( x : \;\cos \left( x\right) = \frac{1}{2} \) | Solution. We have the obvious solution to the equation \( x = {\cos }^{-1}\left( \frac{1}{2}\right) = \frac{\pi }{3} \) . However, since \( \cos \left( {-x}\right) = \cos \left( x\right) \), there is another solution given by taking \( x = - \frac{\pi }{3} \) :\n\n\[ \cos \left( {-\frac{\pi }{3}}\right) = \cos \left( \... | Yes |
Solve for \( x \). \[ \text{a)}\cos \left( x\right) = - \frac{\sqrt{2}}{2} \] | a) First, we need to find \( {\cos }^{-1}\left( {-\frac{\sqrt{2}}{2}}\right) \). From equation (19.3) we know that \( {\cos }^{-1}\left( {-c}\right) = \pi - {\cos }^{-1}\left( c\right) \), so that: \[ {\cos }^{-1}\left( {-\frac{\sqrt{2}}{2}}\right) = \pi - {\cos }^{-1}\left( \frac{\sqrt{2}}{2}\right) = \pi - \frac{\pi ... | Yes |
Solve for \( x : \;\sin \left( x\right) = \frac{\sqrt{2}}{2} \) | Solution. First, we can find one obvious solution \( x = {\sin }^{-1}\left( \frac{\sqrt{2}}{2}\right) = \frac{\pi }{4} \) . Furthermore, from the top right equation in (17.3), we have that \( \sin \left( {\pi - x}\right) = \sin \left( x\right) \) , so that another solution is given by \( \pi - \frac{\pi }{4} \) :\n\n\[... | Yes |
Solve for \( x \). a) \(\sin \left( x\right) = \frac{1}{2}\) | a) First, we calculate \( {\sin }^{-1}\left( \frac{1}{2}\right) = \frac{\pi }{6} \). The general solution is therefore,\n\n\[ x = {\left( -1\right) }^{n} \cdot \frac{\pi }{6} + n \cdot \pi ,\;\text{ where }n = 0, \pm 1, \pm 2, \pm 3,\ldots \] | Yes |
Find the general solution of the equation, and state at least 5 distinct solutions. | \[ \text{a)}\sin \left( x\right) = - \frac{1}{2}\;\text{b)}\cos \left( x\right) = - \frac{\sqrt{3}}{2} \] Solution. a) We already calculated the general solution in example 20.9 (b). It is \[ x = {\left( -1\right) }^{n + 1} \cdot \frac{\pi }{6} + n \cdot \pi ,\;\text{ where }n = 0, \pm 1, \pm 2, \pm 3,\ldots \] We simp... | Yes |
Example 20.11. Solve for \( x \). \[ \text{a)}2\sin \left( x\right) - 1 = 0\;\text{b)}\sec \left( x\right) = - \sqrt{2}\;\text{c)}7\cot \left( x\right) + 3 = 0 \] | Solution. a) Solving for \( \sin \left( x\right) \), we get \[ 2\sin \left( x\right) - 1 = 0\overset{\left( +1\right) }{ \Rightarrow }2\sin \left( x\right) = 1\overset{\left( \div 2\right) }{ \Rightarrow }\sin \left( x\right) = \frac{1}{2} \] One solution of \( \sin \left( x\right) = \frac{1}{2} \) is \( {\sin }^{-1}\l... | Yes |
Example 20.13. Solve the equation with the calculator. Approximate the solution to the nearest thousandth.\n\n\\[ \n\\text{a)}2\\sin \\left( x\\right) = 4\\cos \\left( x\\right) + 3\\;\\text{b)}5\\cos \\left( {2x}\\right) = \\tan \\left( x\\right) \n\\] | Solution. a) We rewrite the equation as \\( 2\\sin \\left( x\\right) - 4\\cos \\left( x\\right) - 3 = 0 \\), and use the calculator to find the graph of the function \\( f\\left( x\\right) = 2\\sin \\left( x\\right) - 4\\cos \\left( x\\right) - 3 \\) . The zeros of the function \\( f \\) are the solutions of the initia... | No |
Example 21.3. Perform the operation.\n\n\[ \text{a)}\left( {2 - {3i}}\right) + \left( {-6 + {4i}}\right) \] | Solution. a) Adding real and imaginary parts, respectively, gives,\n\n\[ \left( {2 - {3i}}\right) + \left( {-6 + {4i}}\right) = 2 - {3i} - 6 + {4i} = - 4 + i. \] | Yes |
Example 21.5. Find the absolute value of the complex numbers below. | Solution. The absolute values are calculated as follows.\n\n\[ \text{a)}\left| {5 - {3i}}\right| = \sqrt{{5}^{2} + {\left( -3\right) }^{2}} = \sqrt{{25} + 9} = \sqrt{34} \]\n\nb) \( \;\left| {-8 + {6i}}\right| = \sqrt{{\left( -8\right) }^{2} + {6}^{2}} = \sqrt{{64} + {36}} = \sqrt{100} = {10} \)\n\nc) \( \;\left| {7i}\... | Yes |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.