## H1011 Quizzes

Quiz 5: Maxima, minima and points of inflection
Question 1 Questions
Let $f\left(x\right)={x}^{3}-3{x}^{2}+7$. Which of the following statements are correct?
 a) The graph of $y=f\left(x\right)$ has only one turning point, at $x=2$. b) The absolute maximum value of $f\left(x\right)$ is $7$. c) The graph of $y=f\left(x\right)$ has a local maximum at $x=0$. d) The graph of $y=f\left(x\right)$ has a local minimum at $x=0$. e) The graph of $y=f\left(x\right)$ has one point of inflection.

There is at least one mistake.
For example, choice (a) should be False.
There is at least one mistake.
For example, choice (b) should be False.
There is a local maximum value of $7$, but it is not the absolute maximum.
There is at least one mistake.
For example, choice (c) should be True.
There is at least one mistake.
For example, choice (d) should be False.
There is a local maximum at $x=0$.
There is at least one mistake.
For example, choice (e) should be True.
Correct!
1. False
2. False There is a local maximum value of $7$, but it is not the absolute maximum.
3. True
4. False There is a local maximum at $x=0$.
5. True
You are given that the function $f\left(x\right)=\frac{2\left(x+3\right)}{{x}^{2}+x-2}$ has an absolute maximum on the interval $-2. Find this maximum.
 a) $-\frac{2}{9}$ b) -10 c) -2 d) -5

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
${f}^{\prime }\left(x\right)=-\frac{2\left(x+5\right)\left(x+1\right)}{\left(x+2\right)\left(x-1\right)}$. So the only critical value of $f$ on the given interval occurs at $x=-1$. Given that $f$ has a maximum it must be at $x=-1$. Moreover, we find that $f\left(-1\right)=-2$.
Note that $f\left(x\right)$ is not defined at $x=-2$ or at $x=1$, but $f\left(x\right)$ approaches $-\infty$ as $x$ approaches 1 from below or $-2$ from above.
Choice (d) is incorrect
Find the absolute minimum of the function $f\left(t\right)=\frac{{t}^{2}+2t}{{e}^{t}}$ on the interval $-2\le t\le 2$.
 a) $\frac{2-2\sqrt{2}}{{e}^{-\sqrt{2}}}$ b) $\sqrt{2}$ c) $\frac{2+2\sqrt{2}}{{e}^{\sqrt{2}}}$ d) $-\sqrt{2}$

Choice (a) is correct!
${f}^{\prime }\left(x\right)=-\frac{2-{t}^{2}}{{e}^{t}}$
Critical points are at $t=±\sqrt{2}$. There is a minimum at $t=-\sqrt{2}$.
$f\left(-\sqrt{2}\right)=\frac{2-2\sqrt{2}}{{e}^{-\sqrt{2}}}\approx -3.41$
$f\left(2\right)=1.08$, $f\left(-2\right)=0$
therefore the absolute minimum is $\frac{2-2\sqrt{2}}{{e}^{-\sqrt{2}}}$ at $t=-\sqrt{2}$.
Choice (b) is incorrect
This is a critical point.
Choice (c) is correct!
This is the absolute maximum.
Choice (d) is incorrect
This is a critical point.
Find the points of inflection of the function $f\left(x\right)={x}^{4}-12{x}^{3}+6x-9$ on the interval $-2\le x\le 10$.
 a) $x=0$, $6$ b) $x=0$, $-6$ c) $x=±\sqrt{6}$ d) $x=±\sqrt{12}$

Choice (a) is correct!
$\begin{array}{rcll}f\left(x\right)& =& {x}^{4}-12{x}^{3}+6x-9& \text{}\\ {f}^{\prime }\left(x\right)& =& 4{x}^{3}-36{x}^{2}+6& \text{}\\ {f}^{″}\left(x\right)& =& 12{x}^{2}-72x& \text{}\end{array}$ ${f}^{″}\left(x\right)=0$ when $12{x}^{2}-72x=0⇒x\left(x-6\right)=0$
So $x=0$, $x=6$.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Find the points of inflection of the function $f\left(x\right)=sin2x+{x}^{2}$ on the interval $0\le x\le \frac{\pi }{2}$.
 a) $0$, $\frac{\pi }{4}$ b) $0$, $\frac{\pi }{2}$ c) $\frac{\pi }{6}$, $\frac{5\pi }{6}$ d) $\frac{\pi }{12}$, $\frac{5\pi }{12}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
$\begin{array}{rcll}f\left(x\right)& =& sin2x+{x}^{2}& \text{}\\ {f}^{\prime }\left(x\right)& =& 2cos2x+2x& \text{}\\ {f}^{″}\left(x\right)& =& -4sin2x+2& \text{}\end{array}$ ${f}^{″}\left(x\right)=0$ when $4\left(\frac{1}{2}-sin2x\right)=0$ i.e when $sin2x=\frac{1}{2}$
i.e $\begin{array}{rcll}2x& =& \frac{\pi }{6},\frac{5\pi }{6}& \text{}\\ x& =& \frac{\pi }{12},\frac{5\pi }{12}.& \text{}\end{array}$
The graph below is of the function $y=f\left(x\right)$.
Which of the following statements gives reasonable values for $x$ which satisfy the conditions.
 a) ${f}^{\prime }\left(x\right)<0$ when $-1${f}^{″}\left(x\right)<0$ when $x>1$${f}^{\prime }\left(x\right)>0$ and ${f}^{″}\left(x\right)=0$ when $x=1$. b) ${f}^{\prime }\left(x\right)<0$ when $-1${f}^{″}\left(x\right)<0$ when $x>1$${f}^{\prime }\left(x\right)>0$ and ${f}^{″}\left(x\right)=0$ for no value of $x$. c) ${f}^{\prime }\left(x\right)<0$ when $x<-1$ and $x>3$${f}^{″}\left(x\right)>0$ when $x<1$${f}^{\prime }\left(x\right)>0$ and ${f}^{″}\left(x\right)=0$ when $x=1$. d) ${f}^{\prime }\left(x\right)<0$ when $x<-1$ and $x>3$${f}^{″}\left(x\right)>0$ when $x<1$${f}^{\prime }\left(x\right)>0$ and ${f}^{″}\left(x\right)=0$ for no value of $x$.

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
Choice (d) is incorrect
The graph below is of the function $y=f\left(x\right)$.
Which of the following statements gives reasonable values for $x$ which satisfy the conditions.
 a) ${f}^{\prime }\left(x\right)<0$ when $x<0$ and $5${f}^{″}\left(x\right)>0$ when $3${f}^{\prime }\left(x\right)>0$ and ${f}^{″}\left(x\right)=0$ when $x=6$. b) ${f}^{\prime }\left(x\right)<0$ when $1 and $x>7$${f}^{″}\left(x\right)>0$ when $3${f}^{\prime }\left(x\right)>0$ and ${f}^{″}\left(x\right)=0$ when $x=6$. c) ${f}^{\prime }\left(x\right)<0$ when $x<1$ and $5${f}^{″}\left(x\right)>0$ when $x<3$ and $x>6$${f}^{\prime }\left(x\right)>0$ and ${f}^{″}\left(x\right)=0$ when $x=3$. d) ${f}^{\prime }\left(x\right)<0$ when $1 and $5${f}^{″}\left(x\right)>0$ when $x<3$ and $x>6$${f}^{\prime }\left(x\right)>0$ and ${f}^{″}\left(x\right)=0$ when $x=3$.

Choice (a) is incorrect
Choice (b) is correct!
Choice (c) is incorrect
Choice (d) is incorrect
Find the value of $a$ such that the function $f\left(x\right)=x{e}^{ax}$ has a critical point at $x=3$. (Give your answer correct to 2 decimal places.)

Correct!
We need ${f}^{\prime }\left(3\right)={e}^{3a}\left(3a+1\right)=0$.
Find ${f}^{\prime }\left(3\right)$, put it equal to zero, and solve for $a$.
How many turning points are there on a graph of $y={e}^{x}+sinx+2x$?
$\frac{dy}{dx}>0$ for all $x$, so there are no turning points.
Note that $\frac{dy}{dx}>0$ for all $x$.
The graph of $P\left(t\right)=\frac{2}{1+{e}^{-t}}$ has exactly one point of inflection. Find the value of $P\left(t\right)$ at this point.
${P}^{″}\left(t\right)=\frac{2{e}^{-t}\left({e}^{-t}-1\right)}{{\left(1+{e}^{-t}\right)}^{3}}$, and the only solution of ${P}^{″}\left(t\right)=0$ is $t=0$. So the point of inflection must occur at $t=0$. Substituting $t=0$ in the formula for $P\left(t\right)$ gives $P\left(0\right)=1$.
${P}^{″}\left(t\right)=\frac{2{e}^{-t}\left({e}^{-t}-1\right)}{{\left(1+{e}^{-t}\right)}^{3}}$