## MATH1111 Quizzes

The Second Derivative Quiz
Web resources available Questions

This quiz tests the work covered in Lecture 11 and corresponds to Section 2.5 of the textbook Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et al.).

There is an applet at http://www.walter-fendt.de/m11e/deriv12.htm which allows you to put in the formula for a function and then it draws the first and second derivative functions.

There is a good summary of the second derivative at 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.

Consider the graph of $y=f\left(x\right)$ below.

Which of the following statements are correct? Exactly one option must be correct)
 a) b) c) d)

Choice (a) is incorrect
Try again, the function is decreasing at $A\phantom{\rule{0.3em}{0ex}}.$
Choice (b) is correct!
The function is decreasing and concave down at $A\phantom{\rule{0.3em}{0ex}},$ so
Choice (c) is incorrect
Try again, the function is decreasing at $A\phantom{\rule{0.3em}{0ex}}.$
Choice (d) is incorrect
Try again, the function is concave down at $A\phantom{\rule{0.3em}{0ex}}.$
Let $P\left(t\right)$ be the number of alien sightings at time $t\phantom{\rule{0.3em}{0ex}}.$ We are told that both ${P}^{\prime }\left(t\right)$ and ${P}^{″}\left(t\right)$ are positive.
Which of the statements below best reflect this? Exactly one option must be correct)
 a) There are more and more alien sightings every year. b) There are more alien sightings this year but less than we expected given the the increases over the past few years. c) There are less alien sightings this year but more than we expected given the the decreases over the past few years. d) There are fewer and fewer alien sightings every year.

Choice (a) is correct!
The number of sightings are increasing so ${P}^{\prime }\left(t\right)>0$ and the rate of increase is increasing so ${P}^{″}\left(t\right)>0\phantom{\rule{0.3em}{0ex}}.$
Choice (b) is incorrect
Try again, since ${P}^{″}\left(t\right)>0$ the rate of sightings must be increasing.
Choice (c) is incorrect
Try again, since ${P}^{\prime }\left(t\right)>0$ the number of sightings must be increasing.
Choice (d) is incorrect
Try again, since ${P}^{\prime }\left(t\right)>0$ the number of sightings must be increasing.
Consider the graph of $y=f\left(x\right)$ below.
At which point on the graph is ${f}^{\prime }\left(x\right)>0$ and ${f}^{″}\left(x\right)<0\phantom{\rule{0.3em}{0ex}}.$
Exactly one option must be correct)
 a) $A$ b) $B$ c) $C$ d) $D$ e) None of the above.

Choice (a) is incorrect
Try again, ${f}^{\prime }\left(x\right)<0$ and ${f}^{″}\left(x\right)>0$ at $A\phantom{\rule{0.3em}{0ex}}.$
Choice (b) is incorrect
Try again, ${f}^{\prime }\left(x\right)>0$ and ${f}^{″}\left(x\right)>0$ at $B\phantom{\rule{0.3em}{0ex}}.$
Choice (c) is correct!
$f$ is increasing and concave down at $C$ so ${f}^{\prime }\left(x\right)>0$ and ${f}^{″}\left(x\right)<0$ at $C\phantom{\rule{0.3em}{0ex}}.$
Choice (d) is incorrect
Try again, ${f}^{\prime }\left(x\right)<0$ and ${f}^{″}\left(x\right)<0$ at $D\phantom{\rule{0.3em}{0ex}}.$
Choice (e) is incorrect
There is a correct answer. You need to find a point where $f$ is increasing and concave down.
A function $f$ is decreasing for $x\ge 2$ and $f\left(2\right)=20\phantom{\rule{0.3em}{0ex}},\phantom{\rule{1em}{0ex}}{f}^{\prime }\left(2\right)=-2$ and ${f}^{″}\left(x\right)>0$ for $x\ge 2\phantom{\rule{0.3em}{0ex}}.$

Which of the following is a possible value for $f\left(4\right)\phantom{\rule{0.3em}{0ex}}?$ Exactly one option must be correct)
 a) $f\left(4\right)=24\phantom{\rule{0.3em}{0ex}}.$ b) $f\left(4\right)=21\phantom{\rule{0.3em}{0ex}}.$ c) $f\left(4\right)=18\phantom{\rule{0.3em}{0ex}}.$ d) $f\left(4\right)=16\phantom{\rule{0.3em}{0ex}}.$

Choice (a) is incorrect
Try again, since $f$ is decreasing $f\left(4\right)<20\phantom{\rule{0.3em}{0ex}}.$
Choice (b) is incorrect
Try again, since $f$ is decreasing $f\left(4\right)<20\phantom{\rule{0.3em}{0ex}}.$
Choice (c) is correct!
Since $f$ is decreasing $f\left(4\right)<20$ and since ${f}^{″}\left(x\right)>0$ the function cannot continue to decrease at the same rate as it was decreasing at  $x=2$  therefore $f\left(4\right)>16\phantom{\rule{0.3em}{0ex}}.$
Choice (d) is incorrect
Try again, since ${f}^{″}\left(x\right)>0$ the function cannot continue to decrease at the same rate as it was decreasing at  $x=2$  therefore $f\left(4\right)>16\phantom{\rule{0.3em}{0ex}}.$