## MATH1111 Quizzes

Using the First and Second Derivative Quiz
Web resources available Questions

This quiz tests the work covered in Lecture 17 and corresponds to Section 4.1 of the textbook Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et al.).

There are more web quizzes at Wiley, select Section 1.

You might want to go back to some of the resources for the second derivative if they were a bit hard at the time. They were http://archives.math.utk.edu/visual.calculus/3/graphing.14/ although some of the language is technical. There are also difficult, but useful quizzes at http://archives.math.utk.edu/visual.calculus/3/graphing.2/index.html and http://archives.math.utk.edu/visual.calculus/3/graphing.3/index.html.

There is some detailed information on this topic with exercises with fully worked solutions at http://www.math.ucdavis.edu/ kouba/CalcOneDIRECTORY/graphingdirectory/Graphing.html.

Consider the function $f\left(t\right)={t}^{3}-12t+8$ It has critical points at $t=-2$ and $t=2$.
Differentiate $f$ and draw a sign diagram of ${f}^{\prime }\left(t\right)$ to decide which one of the following statements is true. Exactly one option must be correct)
 a) ${f}^{\prime }$ changes from positive to negative at $t=-2$ hence there is a local maximum at $\left(-2,12\right)$ b) ${f}^{\prime }$ changes from positive to negative at $t=-2$ hence there is a local minimum at $\left(-2,12\right)$ c) ${f}^{\prime }$ changes from negative to positive at $t=-2$ hence there is a local maximum at $\left(-2,12\right)$ d) ${f}^{\prime }$ changes from negative to positive at $t=-2$ hence there is a local minimum at $\left(-2,12\right)$

Choice (a) is correct!
${f}^{\prime }\left(t\right)=3{x}^{2}-12=3\left(x-2\right)\left(x+2\right)$ so our sign diagram looks like

This shows that ${f}^{\prime }$ changes from positive to negative at $t=-2$ and we can see that means that there is a local maximum at $\left(-2,12\right)$
Choice (b) is incorrect
Try again, changing from positive to negative means that there is a local maximum.

Choice (c) is incorrect
Try again, changing from negative to positive means that there is a local minimum.

Choice (d) is incorrect
Try again, ${f}^{\prime }$ does not change from negative to positive. Choose numbers less than -2 and between -2 and 2 and substitute into ${f}^{\prime }$. For example, ${f}^{\prime }\left(-3\right)=15$ and ${f}^{\prime }\left(0\right)=-12$ so ${f}^{\prime }$ changes from positive to negative.
Which of the following values for $a$ and $b$ give the function $f\left(x\right)={x}^{2}+ax+b$ a local maximum at $\left(2,4\right)$ Exactly one option must be correct)
 a) $a=-4$ and $b=8$ b) $a=4$ and $b=-8$ c) $a=2$ and $b=-4$ d) None of the above.

Choice (a) is incorrect
Try again, there is a local minimum at $\left(2,4\right)$ for the function $f\left(x\right)={x}^{2}-4x+8$
Choice (b) is incorrect
Try again, there is a local minimum at $\left(-2,-12\right)$ for the function $f\left(x\right)={x}^{2}+4x-8$
Choice (c) is incorrect
Try again, perhaps did not differentiate the function correctly.
Choice (d) is correct!
The curve $f\left(x\right)={x}^{2}+ax+b$ is concave up everywhere, since ${f}^{″}\left(x\right)=2\phantom{\rule{0.3em}{0ex}},$ so it cannot have a local maximum. Let’s find the function where $f\left(x\right)={x}^{2}+ax+b$ has a critical value at $\left(2,4\right)$. Since ${f}^{\prime }\left(x\right)=2x+a$ we have ${f}^{\prime }\left(2\right)=4+a=0$, so$a=-4$. Now $f\left(x\right)={x}^{2}-4x+b$ and $f\left(2\right)=4-8+b=4$, so $b=8$. Hence $f\left(x\right)={x}^{2}-4x+8$
Consider the function $y={e}^{{x}^{2}}$ Find the first and second derivatives of $y$ and decide which one of the following statements is correct. Exactly one option must be correct)
 a) There is a local minimum at $\left(0,0\right)$ b) There is a local minimum at $\left(0,1\right)$ c) There is a local maximum at $\left(1,0\right)$ d) There is a local maximum at $\left(\frac{1}{2},{e}^{0.25}\right)$

Choice (a) is incorrect
Try again, at $\left(x,y\right)=\left(0,1\right)$
Choice (b) is correct!
$\frac{dy}{dx}=2x{e}^{{x}^{2}}$ and so there is a critical point at $x=0$: $\begin{array}{llll}\hfill \frac{{d}^{2}y}{d{x}^{2}}& =2x{e}^{{x}^{2}}+2x\left(2x\right){e}^{{x}^{2}}=\left(2+4{x}^{2}\right){e}^{{x}^{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{dy}{dx}|{}_{x=0}& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{{d}^{2}y}{d{x}^{2}}|{}_{x=0}& =2>0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ Therefore there is a local minimum at $\left(0,1\right)$
Choice (c) is incorrect
Try again, at $\left(x,y\right)=\left(0,e\right)$
Choice (d) is incorrect
Try again, $x=\frac{1}{2}$ is not a critical value.
Consider the function $f\left(x\right)={x}^{2}{\left(x-2\right)}^{3}$
Find the first and second derivatives of $f$ and decide which of the following statements are true. (Zero or more options can be correct)
 a) There are stationary points of inflection at $x=0$ and $x=\frac{4}{5}$, and a local minimum at $x=2$ b) There is a stationary point of inflection at $x=2$, a local minimum at $x=\frac{4}{5}$, and a local maximum at $x=0$ c) There are points of inflection at $x=2$, $x=\frac{4+\sqrt{6}}{5}$ and $x=\frac{4-\sqrt{6}}{5}$ d) There are points of inflection at $x=\frac{4}{5}$ and $x=0$ e) There is a local maximum at $\left(0,0\right)$ f) There is only one point of inflection and it is at $x=2$

There is at least one mistake.
For example, choice (a) should be False.
You may have used the function to construct the sign diagram, instead of the derivative.
There is at least one mistake.
For example, choice (b) should be True.
$\begin{array}{llll}\hfill {f}^{\prime }\left(x\right)& =2x{\left(x-2\right)}^{3}+{x}^{2}×\left(3{\left(x-2\right)}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={\left(x-2\right)}^{2}x\left(2\left(x-2\right)+3x\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={\left(x-2\right)}^{2}x\left(5x-4\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ This tells us that there are critical points at $x=0,\frac{4}{5}$ and $2$. Construct a sign diagram to determine the nature of these points: Hence there is a stationary points of inflection at $x=2$ and a local minimum at $x=\frac{4}{5}$ and a local maximum at $x=0$
There is at least one mistake.
For example, choice (c) should be True.
$\begin{array}{llll}\hfill {f}^{\prime }\left(x\right)& ={\left(x-2\right)}^{2}\left(5{x}^{2}-4x\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {f}^{″}\left(x\right)& =2\left(x-2\right)\left(5{x}^{2}-4x\right)+{\left(x-2\right)}^{2}\left(10x-4\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(x-2\right)\left(2\left(5{x}^{2}-4x\right)+\left(x-2\right)\left(10x-4\right)\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =4\left(x-2\right)\left(5{x}^{2}-8x+2\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ Using the quadratic formula to find where $5{x}^{2}-8x+2=0$ we have:
$x=\frac{8±\sqrt{64-40}}{2×5}=\frac{8±\sqrt{24}}{10}=\frac{4±\sqrt{6}}{5}.$
Note that $\sqrt{24}=\sqrt{4×6}=2\sqrt{6}$. So there are points of inflection at $x=2$, $x=\frac{4+\sqrt{6}}{5}$ and $x=\frac{4-\sqrt{6}}{5}$
There is at least one mistake.
For example, choice (d) should be False.
You have made the same mistake as in b).
There is at least one mistake.
For example, choice (e) should be True.
This is true but there are more features.
There is at least one mistake.
For example, choice (f) should be False.
You can factorize the second derivative and get 3 points of inflection.
Correct!
1. False You may have used the function to construct the sign diagram, instead of the derivative.
2. True $\begin{array}{llll}\hfill {f}^{\prime }\left(x\right)& =2x{\left(x-2\right)}^{3}+{x}^{2}×\left(3{\left(x-2\right)}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={\left(x-2\right)}^{2}x\left(2\left(x-2\right)+3x\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={\left(x-2\right)}^{2}x\left(5x-4\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ This tells us that there are critical points at $x=0,\frac{4}{5}$ and $2$. Construct a sign diagram to determine the nature of these points: Hence there is a stationary points of inflection at $x=2$ and a local minimum at $x=\frac{4}{5}$ and a local maximum at $x=0$
3. True $\begin{array}{llll}\hfill {f}^{\prime }\left(x\right)& ={\left(x-2\right)}^{2}\left(5{x}^{2}-4x\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {f}^{″}\left(x\right)& =2\left(x-2\right)\left(5{x}^{2}-4x\right)+{\left(x-2\right)}^{2}\left(10x-4\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\left(x-2\right)\left(2\left(5{x}^{2}-4x\right)+\left(x-2\right)\left(10x-4\right)\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =4\left(x-2\right)\left(5{x}^{2}-8x+2\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$ Using the quadratic formula to find where $5{x}^{2}-8x+2=0$ we have:
$x=\frac{8±\sqrt{64-40}}{2×5}=\frac{8±\sqrt{24}}{10}=\frac{4±\sqrt{6}}{5}.$
Note that $\sqrt{24}=\sqrt{4×6}=2\sqrt{6}$. So there are points of inflection at $x=2$, $x=\frac{4+\sqrt{6}}{5}$ and $x=\frac{4-\sqrt{6}}{5}$
4. False You have made the same mistake as in b).
5. True This is true but there are more features.
6. False You can factorize the second derivative and get 3 points of inflection.