COMPUTER EXPLORATIONSUse a CAS to perform the

Problem 20E In Exercise, find the slope of the curve at the point indicated.

Problem 18E Find the slope of the function’s graph at the given point. Then find an equation for the line tangent to the graph there.

Problem 19E In Exercise, find the slope of the curve at the point indicated.

Problem 26E At what points do the graphs of the functions in exercise have horizontal tangents?

Problem 27E Find equations of all lines having slope –1 that are tangent to the curve

Problem 25E At what points do the graphs of the functions in exercise have horizontal tangents?

Problem 28E Find an equation of the straight line having slope ¼ that is tangent to the curve

Problem 41E Vertical Tangents Graph the curves. a. Where do the graphs appear to have vertical tangents? b. Confirm your findings in part (a) with limit calculations. But before you do, read the introduction to Exercises 35 and 36.

Problem 29E Object dropped from a tower An object is dropped from the top of a 100-m-high tower. Its height above ground after t sec is How fast is it falling 2 sec after it is dropped?

Problem 30E Speed of a rocket At t sec after liftoff, the height of a rocket is 3t 2 ft.How fast is the rocket climbing 10 sec after liftoff ?

Problem 31E Circle’s changing area What is the rate of change of the area of a circle with respect to the radius when the radius is r = 3?

Problem 33E Show that the line y = mx + b is its own tangent line at any point (x0, mx0 + b).

Problem 32E Ball’s changing volume What is the rate of change of the volume of a ball with respect to the radius when the radius is r = 2.

Problem 35E Does the graph of have a tangent at the origin? Give reasons for your answer.

Problem 34E Find the slope of the tangent to the curve at the point where x = 4.

Problem 39E Vertical Tangents Graph the curves. a. Where do the graphs appear to have vertical tangents? b. Confirm your findings in part (a) with limit calculations. But before you do, read the introduction to Exercises 37 and 38. Exercise 37 Does the graph of have a vertical tangent at the origin? Give reasons for your ans Exercise 38 Does the graph of have a vertical tangent at the point (0, 1)? Give reasons for your answer.

Problem 40E Vertical Tangents Graph the curves. a. Where do the graphs appear to have vertical tangents? b. Confirm your findings in part (a) with limit calculations. But before you do, read the introduction to Exercises 35 and 36.

In Exercises 5-10, find an equation for the tangent to the curve at the given point. Then sketch the curve and tangent together. \(y=4-x^{2}\), (-1, 3) Equation Transcription: Text Transcription: y=4-x^2

In Exercises 1–4, use the grid and a straight edge to make a rough estimate of the slope of the curve (in y-units per x-unit) at the points \(P_{1}\) and \(P_{2}\). Equation Transcription: Text Transcription: P_1 P_2

Use a CAS to perform the following steps for the functions in Exercises 49-52 a. Plot \(y=f(x)\) over the interval \(\left(x_{0}-1 / 2\right) \leq x \leq\left(x_{0}+3\right)\). b. Holding \(x_{0}\) fixed, the difference quotient \(q(h)=\frac{f\left(x_{0}+h\right)-f\left(x_{0}\right)}{h}\) at \(x_{0}\) becomes a function of the step size \(h\). Enter this function into your CAS workspace. c. Find the limit of \(q\) as \(h \rightarrow 0\). d. Define the secant lines \(y=f\left(x_{0}\right)+q \cdot\left(x-x_{0}\right)\) for \(h\) = 3, 2, and 1. Graph them together with \(f\) and the tangent line over the interval in part (a). \(f(x)=\cos x+4\sin(2x),\quad\ \ \ x_0=\pi\) Equation Transcription: Text Transcription: y=f(x) (x_0-1/2) leq x leq (x_0+3) x_0 q(h) = f(x_0 + h) - f(x_0) / h x_0 h q h rightarrow 0 y = f(x_0) + q cdot (x - x_0) h f f(x) = cos x + 4 sin(2x), x_0 = pi

In Exercises 1–4, use the grid and a straight edge to make a rough estimate of the slope of the curve (in y-units per x-unit) at the points \(P_{1}\) and \(P_{2}\). Equation Transcription: Text Transcription: P_1 P_2

In Exercises 1–4, use the grid and a straight edge to make a rough estimate of the slope of the curve (in y-units per x-unit) at the points \(P_{1}\) and \(P_{2}\). Equation Transcription: Text Transcription: P_1 P_2

In Exercises 1–4, use the grid and a straight edge to make a rough estimate of the slope of the curve (in y-units per x-unit) at the points \(P_{1}\) and \(P_{2}\). Equation Transcription: Text Transcription: P_1 P_2

In Exercises 5-10, find an equation for the tangent to the curve at the given point. Then sketch the curve and tangent together. \(y=(x-1)^{2}+1\), (1, 1) Equation Transcription: Text Transcription: y = (x-1)^2 + 1

In Exercises 5-10, find an equation for the tangent to the curve at the given point. Then sketch the curve and tangent together. \(y=2 \sqrt{x}\), (1, 2) Equation Transcription: Text Transcription: y = 2 sqrt x

In Exercises 5-10, find an equation for the tangent to the curve at the given point. Then sketch the curve and tangent together. \(y=\frac{1}{x^{2}}\), (-1, 1) Equation Transcription: Text Transcription: y = 1/x^2

In Exercises 5-10, find an equation for the tangent to the curve at the given point. Then sketch the curve and tangent together. \(y=x^{3}\), (-2, -8) Equation Transcription: Text Transcription: y = x^3

In Exercises 11-18, find the slope of the function's graph at the given point. Then find an equation for the line tangent to the graph there. \(f(x)=x^{2}+1\), (2, 5) Equation Transcription: Text Transcription: f(x) = x^2 + 1

In Exercises 11-18, find the slope of the function's graph at the given point. Then find an equation for the line tangent to the graph there. \(f(x)=x-2 x^{2}\), (1, -1) Equation Transcription: Text Transcription: f(x) = x - 2x^2

In Exercises 11-18, find the slope of the function's graph at the given point. Then find an equation for the line tangent to the graph there. \(g(x)=\frac{x}{x-2}\), (3, 3) Equation Transcription: Text Transcription: g(x) = x / x-2

In Exercises 5-10, find an equation for the tangent to the curve at the given point. Then sketch the curve and tangent together. \(y=\frac{1}{x^3},\quad\ \ \ \left(-2,-\frac{1}{8}\right)\) Equation Transcription: Text Transcription: y = 1/x^3, (-2, -1/8)

In Exercises 11-18, find the slope of the function's graph at the given point. Then find an equation for the line tangent to the graph there. \(g(x)=\frac{8}{x^{2}}\), (2, 2) Equation Transcription: Text Transcription: g(x) = 8/x^2

In Exercises 11-18, find the slope of the function's graph at the given point. Then find an equation for the line tangent to the graph there. \(h(t)=t^{3}\), (2, 8) Equation Transcription: Text Transcription: h(t)=t^3

In Exercises 11-18, find the slope of the function's graph at the given point. Then find an equation for the line tangent to the graph there. \(h(t)=t^{3}+3t\), (1, 4) Equation Transcription: Text Transcription: h(t)=t^3 + 3t

In Exercises 11-18, find the slope of the function's graph at the given point. Then find an equation for the line tangent to the graph there. \(f(x)=\sqrt{x}\), (4, 2) Equation Transcription: Text Transcription: f(x) = sqrt x

In Exercises 19-22, find the slope of the curve at the point indicated. \(y=\frac{1}{x-1},\quad\ \ \ x=3\) Equation Transcription: Text Transcription: y = 1/x-1, x=3

In Exercises 19-22, find the slope of the curve at the point indicated. \(y=\frac{x-1}{x+1},\quad\ \ \ x=0\) Equation Transcription: Text Transcription: y = x-1/x+1, x=0

Growth of yeast cells In a controlled laboratory experiment, yeast cells are grown in an automated cell culture system that counts the number \(P\) of cells present at hourly intervals. The number after \(t\) hours is shown in the accompanying figure. a. Explain what is meant by the derivative \(P^{\prime}(5)\). What are its units? b. Which is larger, \(P^{\prime}(2)\) or \(P^{\prime}(3)\)? Give a reason for your answer. c. The quadratic curve capturing the trend of the data points (see Section 1.4) is given by \(P(t)=6.10 t^{2}-9.28 t+16.43\). Find the instantaneous rate of growth when \(t=5\) hours. Equation Transcription: Text Transcription: P t P'(5) P'(2) P'(3) P(t)=6.10t^2 - 9.28t + 16.43 t=5

Effectiveness of a drug On a scale from 0 to 1, the effectiveness \(E\) of a pain-killing drug \(t\) hours after entering the bloodstream is displayed in the accompanying figure. a. At what times does the effectiveness appear to be increasing? What is true about the derivative at those times? b. At what time would you estimate that the drug reaches its maximum effectiveness? What is true about the derivative at that time? What is true about the derivative as time increases in the 1 hour before your estimated time? Equation Transcription: Text Transcription: E t

Vertical Tangents We say that a continuous curve \(y=f(x)\) has a vertical tangent at the point where \(x=x_{0}\) if the limit of the difference quotient is \(\infty\) or \(-\infty\). For example, \(y=x^{1 / 3}\) has a vertical tangent at \(x=0\) (see accompanying figure): \(\lim _{h \rightarrow 0} \frac{f(0+h)-g(0)}{h}=\lim _{h \rightarrow 0} \frac{h^{1 / 3}-0}{h}=\lim _{h \rightarrow 0} \frac{1}{h^{2 / 3}}=\infty\) VERTICAL TANGENT AT ORIGIN However, \(y=x^{2 / 3}\) has \(no\) vertical tangent at \(x=0\) (see next figure): \(\lim _{h \rightarrow 0} \frac{g(0+h)-g(0)}{h}=\lim _{h \rightarrow 0} \frac{h^{2 / 3}-0}{h}=\lim _{h \rightarrow 0} \frac{1}{h^{1 / 3}}\) does not exist, because the limit is \(\infty\) from the right and \(-\infty\) from the left. NO VERTICAL TANGENT AT ORIGIN Does the graph of \(U(x)=\left\{\begin{array}{ll}0, & x<0 \\1, & x \geq 0\end{array}\right.\) have a vertical tangent at the point (0,1)? Give reasons for your answer. Equation Transcription: { Text Transcription: y=f(x) x=x_0 infty -infty y=x^1/3 x=0 lim over h rightarrow 0 f(0+h)-f(0)/h=lim over h rightarrow 0 h^1/3-0/h =lim over h rightarrow 0 1/h^2/3=infty y=x^2/3 no x=0 lim over h rightarrow 0 g(0+h)-g(0)/h lim over h rightarrow 0 h2/3-0/h =lim over h rightarrow 0 1/h^1/3 infty -infty U(x)={_1, leq x 0 ^0, x < 0

Does the graph of \(g(x)=\left\{\begin{array}{ll} x \sin (1 / x), & x \neq 0 \\ 0, & x=0 \end{array}\right.\) have a tangent at the origin? Give reasons for your answer. Equation Transcription: Text Transcription: g(x) = {_0, x=0 ^x sin (1/x), x neq 0

We say that a continuous curve \(y=f(x)\) has a vertical tangent at the point where \(x=x_{0}\) if the limit of the difference quotient is \(\infty\) or \(-\infty\). For example, \(y=x^{1 / 3}\) has a vertical tangent at \(x=0\) (see accompanying figure): \(\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{h^{1 / 3}-0}{h}=\lim _{h \rightarrow 0} \frac{1}{h^{2 / 3}}=\infty\). VERTICAL TANGENT AT ORIGIN However, \(y=x^{2 / 3}\) has no vertical tangent at \(x=0\) (see next figure): \(\lim _{h \rightarrow 0} \frac{g(0+h)-g(0)}{h}=\lim _{h \rightarrow 0} \frac{h^{2 / 3}-0}{h}=\lim _{h \rightarrow 0} \frac{1}{h^{1 / 3}}\) does not exist, because the limit is \(\infty\) from the right and \(-\infty\) from the left. NO VERTICAL TANGENT AT ORIGIN Does the graph of \(f(x)=\left\{\begin{array}{rr} -1, & x<0 \\ 0, & x=0 \\ 1, & x>0 \end{array}\right.\) have a vertical tangent at the origin? Give reasons for your answer. Equation Transcription: { Text Transcription: y=f(x) x=x_0 infty - infty y=x^1/3 x=0 lim_h rightarrow 0 f(0+h)-g(0)/h = lim_h rightarrow 0 h^1/3 -0/h = lim_h rightarrow 0 1/h^2/3 = infty. y=x^2/3 x=0 lim_h rightarrow 0 g(0+h)-g(0)/h = lim_h rightarrow 0 h^2/3 -0/h = lim_h rightarrow 0 1/h^? infty - infty f(x) = { _1, x>0 ^0, x=0 ^^-1, x<0

Graph the curves in Exercises 39-48. a. Where do the graphs appear to have vertical tangents? b. Confirm your findings in part (a) with limit calculations. But before you do, read the introduction to Exercises 37 and 38 . \(y=x^{3 / 5}\) Equation Transcription: Text Transcription: y=x^3/5

Graph the curves in Exercises 39-48. a. Where do the graphs appear to have vertical tangents? b. Confirm your findings in part (a) with limit calculations. But before you do, read the introduction to Exercises 37 and 38 . \(y=4 x^{2 / 5}-2 x\) Equation Transcription: Text Transcription: y=4x^2/5 - 2x

Graph the curves in Exercises 39-48. a. Where do the graphs appear to have vertical tangents? b. Confirm your findings in part (a) with limit calculations. But before you do, read the introduction to Exercises 37 and 38 . \(y=x^{1 / 3}+(x-1)^{1 / 3}\) Equation Transcription: Text Transcription: y=x^1/3 + (x-1)^1/3

Graph the curves in Exercises 39-48. a. Where do the graphs appear to have vertical tangents? b. Confirm your findings in part (a) with limit calculations. But before you do, read the introduction to Exercises 37 and 38 . \(y=x^{2 / 3}-(x-1)^{1 / 3}\) Equation Transcription: Text Transcription: y=x^2/3 - (x-1)^1/3

Graph the curves in Exercises 39-48. a. Where do the graphs appear to have vertical tangents? b. Confirm your findings in part (a) with limit calculations. But before you do, read the introduction to Exercises 37 and 38 . \(y=\left\{\begin{array}{ll} -\sqrt{|x|}, & x \leq 0 \\ \sqrt{x}, & x>0 \end{array}\right.\) Equation Transcription: Text Transcription: y = {_sqrt x, x>0 ^- sqrt|x|, x leq 0

Graph the curves in Exercises 39-48. a. Where do the graphs appear to have vertical tangents? b. Confirm your findings in part (a) with limit calculations. But before you do, read the introduction to Exercises 37 and 38 . \(y=x^{5 / 3}-5 x^{2 / 3}\) Equation Transcription: Text Transcription: y=x^5/3 - 5x^2/3

Graph the curves in Exercises 39-48. a. Where do the graphs appear to have vertical tangents? b. Confirm your findings in part (a) with limit calculations. But before you do, read the introduction to Exercises 37 and 38 . \(y=\sqrt{|4-x|}\) Equation Transcription: Text Transcription: y = sqrt |4-x|

Use a CAS to perform the following steps for the functions in Exercises 49-52. a. Plot \(y=f(x)\) over the interval \(\left(x_{0}-1 / 2\right) \leq x \leq\left(x_{0}+3\right)\). b. Holding \(x_{0}\) fixed, the difference quotient \(q(h)=\frac{f\left(x_{0}+h\right)-f\left(x_{0}\right)}{h}\) at \(x_{0}\) becomes a function of the step size \(h\). Enter this function into your CAS workspace. c. Find the limit of \(q\) as \(h \rightarrow 0\). d. Define the secant lines \(y=f\left(x_{0}\right)+q \cdot\left(x-x_{0}\right)\) for \(h=\) 3, 2, and 1 . Graph them together with \(f\) and the tangent line over the interval in part (a). \(f(x)=x^3+2x,\quad\ \ \ x_0=0\) Equation Transcription: Text Transcription: y=f(x) (x_0 - 1/2) leq x leq (x_0 + 3) x_0 q(h) = f(x_0 + h) - f(x_0) / h x_0 h q h rightarrow 0 y = f(x_0) + q cdot (x - x_0) h= f f(x) = x^3 + 2x, x_0 = 0

Use a CAS to perform the following steps for the functions in Exercises 49-52. a. Plot \(y=f(x)\) over the interval \(\left(x_{0}-1 / 2\right) \leq x \leq\left(x_{0}+3\right)\). b. Holding \(x_{0}\) fixed, the difference quotient \(q(h)=\frac{f\left(x_{0}+h\right)-f\left(x_{0}\right)}{h}\) at \(x_{0}\) becomes a function of the step size \(h\). Enter this function into your CAS workspace. c. Find the limit of \(q\) as \(h \rightarrow 0\). d. Define the secant lines \(y=f\left(x_{0}\right)+q \cdot\left(x-x_{0}\right)\) for \(h=\) 3, 2, and 1 . Graph them together with \(f\) and the tangent line over the interval in part (a). \(f(x)=x+\frac{5}{x},\quad\ \ \ x_0=1\) Equation Transcription: Text Transcription: y=f(x) (x_0 - 1/2) leq x leq (x_0 + 3) x_0 q(h) = f(x_0 + h) - f(x_0) / h x_0 h q h rightarrow 0 y = f(x_0) + q cdot (x - x_0) h= f f(x) = x + 5/x, x_0 = 1

Use a CAS to perform the following steps for the functions in Exercises 49-52. a. Plot \(y=f(x)\) over the interval \(\left(x_{0}-1 / 2\right) \leq x \leq\left(x_{0}+3\right)\). b. Holding \(x_{0}\) fixed, the difference quotient \(q(h)=\frac{f\left(x_{0}+h\right)-f\left(x_{0}\right)}{h}\) at \(x_{0}\) becomes a function of the step size \(h\). Enter this function into your CAS workspace. c. Find the limit of \(q\) as \(h \rightarrow 0\). d. Define the secant lines \(y=f\left(x_{0}\right)+q \cdot\left(x-x_{0}\right)\) for \(h=\) 3, 2, and 1 . Graph them together with \(f\) and the tangent line over the interval in part (a). \(f(x)=x+\sin(2x),\quad\ \ \ x_0=\pi/2\) Equation Transcription: Text Transcription: y=f(x) (x_0 - 1/2) leq x leq (x_0 + 3) x_0 q(h) = f(x_0 + h) - f(x_0) / h x_0 h q h rightarrow 0 y = f(x_0) + q cdot (x - x_0) h= f f(x) = x + sin?(2x), x_0 = pi/2