 ## H1011 Quizzes

Quiz 6: One variable optimisation
Question 1 Questions
Find the relative growth rate of the function $P\left(t\right)=10{e}^{0.3t}$.

Correct!
The relative growth rate is $\frac{{P}^{\prime }\left(t\right)}{P\left(t\right)}$.
Incorrect. Please try again.
Remember that the relative growth rate is $\frac{{P}^{\prime }\left(t\right)}{P\left(t\right)}$.
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
Find the absolute maximum of the function $f\left(x\right)=\frac{x+3}{{x}^{2}+x-2}$ on the interval $-2\le x\le 1$.
 a) $-\frac{2}{18}$ b) -10 c) -1 d) -5

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
${f}^{\prime }\left(x\right)=-\frac{\left(x+5\right)\left(x+1\right)}{\left(x+2\right)\left(x-1\right)}$
So the critical value of $f$ on the given interval is at $x=-1$.
Now $f\left(-1\right)=-1$ and $f\left(-2\right)=f\left(1\right)=-\infty$ so the absolute maximum of the function is -1.
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 $x<1$ 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
How many turning points are there on a graph of $y={e}^{x}+sinx+2x$?

Correct!
$\frac{dy}{dx}>0$ for all $x$, so there are no turning points.
Incorrect. Please try again.
Note that $\frac{dy}{dx}>0$ for all $x$.
Find the value of $P$ at which the graph of $P\left(t\right)=\frac{2}{1+{e}^{-t}}$ has a point of inflection.

Correct!
${P}^{″}\left(t\right)=0$ when $t=0$.
Incorrect. Please try again.
${P}^{″}\left(t\right)=\frac{2{e}^{-t}\left({e}^{-t}-1\right)}{{\left(1+{e}^{-t}\right)}^{3}}$