A Collection of Problems in Differential Calculus: Problems from Calculus I Final Examinations, 2000-2020 Department of Mathematics, Simon Fraser University

Section 6.7 Curve Sketching

1. (a) $f(x)=x^3\text{.}$ (b) $f(x)=x-\mbox{sign} (x)\text{,}$ $x\not= 0\text{.}$ (c) 4. (d) $a=2\sqrt{e}\text{,}$ $b=0.125\text{.}$

2. Local maximum.

3. Observe that $f^\prime(x)>0$ for all $x\in\mathbb{R}\text{.}$

4. (a) Velocity — bottom left; acceleration — bottom right. (b) Speeding up on $(0,1.5)$ and $(4,5)\text{.}$ (c) Approximately 5.8 units.

5. (a) $(3,f(3))\text{.}$ (b) $(-1,f(-1))\text{.}$ (c) $(-0.5,f(-0.5))\text{,}$ $(0.5,f(0.5))\text{,}$ $(2.3,f(2.3))\text{.}$ (d) $(-\infty,-1)$ and $(1.5,\infty)\text{.}$ (e) $(-\infty,0)$ and $(1.5,\infty)\text{.}$ (f) $5\text{.}$ Observe that the second derivative has three zeros.

6. 1. From $f'(x)=12x^2(x-2)$ we conclude that $f'(x)>0$ for $x>2$ and $f'(x)\lt 0$ for $x\lt 2\text{.}$ So $f$ is increasing on $(2,\infty)$ and decreasing on $(-\infty,2)\text{.}$

   2. From $f"(x)=12x(3x-4)$ it follows that $f"(x)>0$ for $x\lt 0$ or $\ds x>\frac{4}{3}$ and $f"(x)\lt 0$ for $\ds x\in \left( 0,\frac{4}{3}\right)\text{.}$ Also $f"(x)=0$ for $x=0$ and $\ds x=\frac{4}{3}\text{.}$ Thus $f$ is concave upward on $(-\infty,0)$ and on $\ds \left( \frac{4}{3},\infty \right)$ and concave downward on $\ds \left( 0,\frac{4}{3}\right)$

   3. Critical numbers are $x=0$ and $x=2\text{.}$ Since $f'(x)$ does not change sign at $x=0$ there is no local maximum or minimum there. (Note also that $f"(0)=0$ and that the second derivative test is inconclusive.) Since $f'(x)$ changes from negative to positive at $x=2$ there is a local minimum at $x=2\text{.}$ (Note also that $f"(2)>0\text{,}$ so second derivative test says there is a local minimum.)

   4. Inflection points are $(0,10)$ and $\ds \left( \frac{4}{3},f\left( \frac{4}{3}\right) \right)\text{.}$

   5. $\ds \lim _{x\to \pm \infty}f(x)= \infty\text{.}$

   For the graph see Figure 6.7.1.

   

7. (a) From $x^2-9>0$ it follows that the domain of the function $f$ is the set $(-\infty,-3)\cup(3,\infty)\text{.}$ (b) The function is not defined at $x=0\text{,}$ so there is no the $y$-intercept. Note that $f(x)\not= 0$ for all $x$ in the domain of $f\text{.}$ (c) From $\ds \lim _{x\to \infty }f(x)=1$ and $\ds \lim _{x\to -\infty }f(x)=-1$ we conclude that there are two horizontal asymptotes, $y=1$ (when $x\to \infty$) and $y=-1$ (when $x\to -\infty$). From $\ds \lim_{x\rightarrow 3^+}f(x)=0$ and $\ds \lim_{x\rightarrow -3^-}f(x)=-\infty$ it follows that there is a vertical asymptote at $x=-3\text{.}$ (d) Since, for all $x$ in the domain of $f\text{,}$ $\displaystyle f'(x) = \frac{3(x-3)}{(x^2-9)^{3/2}}\not= 0$ we conclude that there is no critical number for the function $f\text{.}$ (e) Note that $f'(x)>0$ for $x>3$ and $f'(x)\lt 0$ for $x\lt -3\text{.}$ Thus $f$ increasing on $(3,\infty )$ and decreasing on $(-\infty ,-3)\text{.}$ (f) Since the domain of $f$ is the union of two open intervals and since the function is monotone on each of those intervals, it follows that the function $f$ has neither (local or absolute) a maximum nor a minimum. (g) From $\displaystyle f''(x) =-\frac{6(x-3)(x-\frac{3}{2})}{(x^2-9)^{5/2}}$ it follows that $f''(x)\lt 0$ for all $x$ in the domain of $f\text{.}$ Therefore $f(x)$ is concave downwards on its domain. For the graph see Figure 6.7.2.

8. (a) The domain of the function $f$ is the set $\mathbb{R}\backslash \{ 0\}\text{.}$ The $x$-intercepts are $\pm 1\text{.}$ Since 0 not in domain of $f$ there is no $y$-intercept. (b) From $\displaystyle \lim_{x\rightarrow 0^-} f(x) = -\infty$ and $\ds \lim_{x\rightarrow 0^+} f(x) =\infty$ it follows that the vertical asymptote is the line $x=0\text{.}$ Since $\displaystyle \lim_{x\rightarrow\pm\infty}f(x) = \lim_{x\rightarrow\pm\infty}\left( x-\frac{1}{x}\right)= \pm\infty$ we conclude that there is no horizontal asymptote. Finally, the fact $\displaystyle f(x) = x -\frac{1}{x}$ implies that $f$ has the slant (oblique) asymptote $y=x\text{.}$ (c) For all $x\in \mathbb{R}\backslash \{ 0\}\text{,}$ $\displaystyle f'(x) = \frac{x^2+1}{x^2} >0$ so the function $f$ is increasing on $(-\infty ,0)$ and on $(0,\infty)\text{.}$ The function $f$ has no critical numbers and thus cannot have a local maximum or minimum. (d) Since $\displaystyle f''(x) = -\frac{2}{x^3}$ it follows that $f''(x)>0$ for $x\lt 0$ and $f''(x)\lt 0$ for $x>0\text{.}$ Therefore $f$ is concave upward on $(-\infty ,0)$ and concave downward on $(0,\infty)\text{.}$ There are no points of inflection. (e) See Figure 6.7.2.

   

   

9. See Figure 6.7.3.

10. See Figure 6.7.3.

    

    

11. See Figure 6.7.4.

    

12. See Figure 6.7.5.

13. $f^\prime(x) =3x^2-4x-1\text{,}$ $f''(x) = 6x-4\text{.}$ See Figure 6.7.5.

    

    

14. See Figure 6.7.6.

15. See Figure 6.7.6.

    

    

16. See Figure 6.7.7.

17. See Figure 6.7.7.

    

    

18. Note that the domain of the given function is the set of all real numbers. The $y$-intercept is the point $(0,0)$ and the $x$-intercepts are $( -4,0)$ and $(0,0)\text{.}$ From $\ds y'= \frac{4} {3}x^{1/3}\left( \frac{1}{x}+1\right)$ we conclude that $y'$ is not defined at $x=0$ and that $y'=0$ if $x=-1\text{.}$ Thus the critical numbers are $x=-1$ and $x=0\text{.}$ Also $y'\lt 0$ on $(-\infty,-1)$ and $y'>0$ on $(-1,0)\cup (0,\infty)\text{.}$ Hence the function has a local minimum at $x=-1\text{.}$ Note that the $y$-axis is a vertical asymptote to the graph of the given function. From $\ds y''= \frac{4}{9}x^{-5/3}(x-2)$ it follows that $y''(x)>0$ on $(-\infty,0)\cup (2,\infty$ and $y''(x)\lt 0$ on $(0,2)\text{.}$ Points of inflection are $(0,0)$ and $\ds (2,6\cdot 2^{1/3})\text{.}$ See Fig 6.7.8.

19. Note that the given function is a product of a power function $y=x^{2/3}$ and a linear function $\ds y=\frac{5}{2}-x$ that are both continuous on $\mathbb{R}\text{.}$ See Fig 6.7.8.

    

    

20. The domain is the interval $(0,\infty)\text{.}$ Note that $\ds \lim_{x\rightarrow 0^+}x^x = 1$ and $\ds \lim_{x\rightarrow \infty }x^x = \infty\text{.}$ From $y' = x^x(\ln x + 1)$ we get that the critical number is $\ds x=e^{-1}\text{.}$ By the first derivative test there is a local minimum there. Also, $y'' = x^x[(\ln x+1)^2 + x^{-1}]\text{.}$ See Figure 6.7.9.

21. See Figure 6.7.9.

    

    

22. See Figure 6.7.10.

23. $\ds \lim_{x\rightarrow 0^-}f(x) = 0\text{,}$ $\ds \lim_{x\rightarrow 0^+}f(x)=\infty\text{,}$ $\ds \lim_{x\rightarrow \pm \infty}f(x) = 1\text{.}$ See Figure 6.7.10.

    

    

24. (a) $y=0\text{.}$ (b) Increasing on $(-\infty ,0)\text{.}$ (c) Local maximum at $x=0\text{.}$ (d) Concave down on $(-2,2)\text{.}$ (e) Inflection points at $x=\pm 2\text{.}$

25. (a) $(0,0)\text{,}$ $(3,9e^2)\text{;}$ (b) Increasing on $(-\infty ,3)$ and decreasing on $(3,\infty )\text{.}$ A local (global) maximum at $(3,9e^2)\text{.}$ The other critical point is neither a local maximum nor a local minimum. (c) Note that $x^2-6x+6=(x-(3-\sqrt{3}))(x-(3+\sqrt{3}))\text{.}$ The function is concave up on $(0,3-\sqrt{3})$ and $(3+\sqrt{3},\infty)$ and concave down on $(-\infty ,0)$ and $(3-\sqrt{3},3+\sqrt{3})\text{.}$ The inflection points are $(0,0)\text{,}$ $(3-\sqrt{3}, (3-\sqrt{3})^3e^{2+\sqrt{3}})\text{,}$ and $(3+\sqrt{3},(3+\sqrt{3})^3e^{2-\sqrt{3}})\text{.}$ (d) $\ds \lim _{x\to -\infty }f(x)=-\infty\text{,}$ $\ds \lim _{x\to \infty }f(x)=0\text{.}$ (e) See Figure 6.7.11.

26. See Figure 6.7.11.

    

    

27. See Figure 6.7.12.

28. $\mbox{dom} (f)=\mathbb{R}\text{,}$ $\mbox{dom} (f^\prime)=\mathbb{R}\backslash \{ 0\}\text{.}$ See Figure 6.7.12.

    

    

29. See Figure 6.7.13.

30. See Figure 6.7.13

    

    

31. It is given that the $y$-intercept is the point $(0,-3)\text{.}$ Note that the given function has a vertical asymptote $x=3$ and two horizontal asymptotes, $y=-1\text{,}$ when $x\to -\infty\text{,}$ and $y=2\text{,}$ when $x\to \infty\text{.}$ Also, the function $f$ is decreasing on $(-\infty ,3)$ and $(3,\infty )\text{.}$ Finally, $f$ is concave upwards on $(3,\infty )$ and concave downwards on $(-\infty ,3)\text{.}$ See Figure 6.7.14.

32. (a) The graph has a vertical asymptote $y=0$ and a horizontal asymptote $x=-2\text{.}$ The following table summarizes the rest of the given information.

    $$\begin{equation*} \begin{array}{c||c|c|c|c|c|} \text{Interval} & (-4,-1) & (-1,0) & (0,2) & (2,4) & (4,\infty) \\ \hline \hline \text{Monotonity} & \text{Decreasing} & \text{Increasing} & \text{Increasing}& \text{Decreasing} & \text{Decreasing}\\ \hline \text{Concavity} & \text{Downwards} & \text{Upwards} & \text{Downwards}& \text{Downwards} & \text{Upwards}\\ \hline \end{array} \end{equation*}$$

    (b) There are two points of inflection, $x=-1$ and $x=4\text{.}$ We note that $x=-1$ is also a critical number and that by the first derivative test there is a local minimum at $x=-1\text{.}$ If $f"(-1)=0\text{,}$ then $f'(-1)$ exists and $f^\prime(-1)=0\text{.}$ This would imply that at this point the graph of $f$ is above the tangent line at $x=-1$ which contradicts the fact that the curve crosses its tangent line at each inflection point. It follows that $f'(-1)$ does not exist and therefore $f"(-1)$ does not exist. For a graph see Figure 6.7.14.

    

    

33. (a) $\surd\text{:}$ $r\text{,}$ $s\text{,}$ $g\text{;}$ (b) $\surd\text{:}$ $r\text{,}$ $s\text{,}$ $f\text{,}$ $g\text{;}$ (c) $\surd\text{:}$ $r\text{;}$ (d) $\surd\text{:}$ $g\text{;}$ (e) $\surd\text{:}$ $r\text{.}$

34. (a) $\surd\text{:}$ C, D; (b) $\surd\text{:}$ A; (c) $\surd\text{:}$ A, D; (d) $\surd\text{:}$ A, B, C, D; (e) $\surd\text{:}$ B.

35. $a=-3\text{,}$ $b=7\text{.}$ Solve the system $y(1)=a+b+2=6$ and $y"(1)=6+2a=0\text{.}$