## MATH1901 Quizzes

Quiz 7: Differentiable functions
Question 1 Questions
Find the equation of the tangent line to $y=\sqrt{{e}^{5x}+3}$ at $x=0$. Exactly one option must be correct)
 a) $x-4y+8=0$ b) $x-4y-8=0$ c) $5x-4y+8=0$ d) $5x-4y+4\sqrt{3}=0$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
Choice (d) is incorrect
Find any critical points of the function $f\left(x\right)=sinx+x$ on the interval $\left[-2\pi ,2\pi \right]$, and classify them. Exactly one option must be correct)
 a) The only critical point is a point of inflection at $\left(\pi ,\pi \right)$. b) The only critical point is a minimum point at $\left(\pi ,\pi \right)$. c) There are points of inflection at $\left(-\pi ,-\pi \right)$ and $\left(\pi ,\pi \right)$. d) There are minimum points at $\left(-\pi ,-\pi \right)$ and $\left(\pi ,\pi \right)$. e) There are no critical points.

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
Choice (d) is incorrect
Choice (e) is incorrect
Find the absolute minimum and maximum values of the function $f\left(x\right)=2{x}^{3}-3{x}^{2}-12x+45$  on the closed interval $\left[-3,3\right]$. Exactly one option must be correct)
 a) The absolute minimum is 0, and the absolute maximum is 52. b) The absolute minimum is 0, and the absolute maximum is 25. c) The absolute minimum is 0, and the absolute maximum is 36. d) The absolute minimum is 25, and the absolute maximum is 52. e) The absolute minimum is 36, and the absolute maximum is 52.

Choice (a) is correct!
$\frac{df\left(x\right)}{dx}=6{x}^{2}-6x-12=0$ when $x=2,-1$. $f\left(-1\right)=52,\phantom{\rule{1em}{0ex}}f\left(2\right)=25$ and checking end points $f\left(-3\right)=0,\phantom{\rule{1em}{0ex}}f\left(3\right)=36$. So minimum is $f\left(-3\right)=0$ and maximum is $f\left(-1\right)=52$.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is incorrect
Find the minimum and maximum values of $f\left(x\right)={x}^{3}-9x+8$ on the interval $\left[-3,1\right]$, if they exist. Exactly one option must be correct)
 a) $0$ and $8$ b) $0$ and $8+6\sqrt{3}$ c) $\sqrt{3}$, and $8+6\sqrt{3}$ d) $8-6\sqrt{3}$ and $8+6\sqrt{3}$ e) $8-\sqrt{3}$ and $8+\sqrt{3}$

Choice (a) is incorrect
Choice (b) is correct!
$\frac{df\left(x\right)}{dx}=3{x}^{2}-9=0$ when $x=-\sqrt{3}$ or $\sqrt{3}$. However $\sqrt{3}>1$ and so $x=\sqrt{3}$ is not in the interval $\left[-3,1\right]$. Comparing $f\left(-3\right)=0,\phantom{\rule{1em}{0ex}}f\left(1\right)=-8$ and $f\left(-\sqrt{3}\right)=8+6\sqrt{3}$ gives the correct answer.
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is incorrect
Consider the function $f\left(x\right)=|2x-4|$. Which of the following are true statements? (Zero or more options can be correct)
 a) $f$ is nowhere differentiable. b) $f$ is differentiable except at $x=2$. c) $f$ is a discontinuous function. d) $\frac{df\left(x\right)}{dx}=2$ for all $x\ne 2$. e) $\frac{df\left(x\right)}{dx}=-2$ for all $x<2$.

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 True.
There is at least one mistake.
For example, choice (c) should be False.
There is at least one mistake.
For example, choice (d) should be False.
There is at least one mistake.
For example, choice (e) should be True.
Correct!
1. False
2. True
3. False
4. False
5. True
According to the Mean Value Theorem, what is the largest number of real roots that the equation
${x}^{7}+8x+13=0$
can have? Exactly one option must be correct)
 a) 4 b) 1 c) 0 d) 2 e) 6

Choice (a) is incorrect
Choice (b) is correct!
Let $f\left(x\right)={x}^{7}+8x+13$. Then ${f}^{\prime }\left(x\right)=7{x}^{6}+8$, so ${f}^{\prime }\left(x\right)>0$ for all $x$. Since $li{m}_{x\to ±\infty }f\left(x\right)=±\infty$ the function $f\left(x\right)$ must have at least one root. If $f\left(x\right)$ has two roots, say $a, then $f\left(a\right)=f\left(b\right)=0$. Applying the Mean Value Theorem of $f\left(x\right)$ on the interval $\left[a,b\right]$ we see that there must exist a number $c\in \left[a,b\right]$ such that
${f}^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}=\frac{0-0}{b-a}=0.$
However, we have already seen that ${f}^{\prime }\left(x\right)>0$ for all $x$, so this is not possible. Hence, $f\left(x\right)$ has exactly one root.
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is incorrect
Which of the following statements are true : (Zero or more options can be correct)
 a) If ${f}^{\prime }\left(c\right)=0$, then $f\left(x\right)$ has a maximum or minimum value at $x=c$. b) If $\frac{df\left(x\right)}{dx}=\frac{dg\left(x\right)}{dx}$ for all $x$ in an open interval $I$, then $f\left(x\right)=g\left(x\right)$ on $I$. c) If $f\left(x\right)$ is differentiable on the open interval $\left(a,b\right)$, and $c$ is a point of local maximum for $f$ in $\left(a,b\right)$, then ${f}^{\prime }\left(c\right)=0$.

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 at least one mistake.
For example, choice (c) should be True.
Correct!
1. False
2. False
3. True
If $f\left(x\right)=2{x}^{3}+{x}^{2}-x-1$, since $f\left(x\right)$ is differentiable for all $x$, by the mean value theorem, there exists a $c$ in [0,2] such that
${f}^{\prime }\left(c\right)=\frac{f\left(2\right)-f\left(0\right)}{2}.$
Find all possible values of $c$. Exactly one option must be correct)
 a) $c=1$ b) $c=\frac{-1±\sqrt{61}}{6}$ c) $c=±1$ d) $c=\frac{-1-\sqrt{61}}{6}$ e) $c=\frac{-1+\sqrt{61}}{6}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is correct!
Note that $\frac{-1-\sqrt{61}}{6}$ is not in the interval $\left[0,2\right]$.
Let $f\left(x\right)=|x-1|$. Then it is easy to see that there is no $c\in \left[0,3\right]$ satisfying
${f}^{\prime }\left(c\right)=\frac{f\left(3\right)-f\left(0\right)}{3}.$
This does not contradict the mean value theorem because (tick each correct answer): (Zero or more options can be correct)
 a) $f\left(x\right)$ is not continuous on $\left[0,3\right]$. b) $f\left(x\right)$ is not differentiable on $\left(0,3\right)$. c) $f\left(x\right)$ is not continuous everywhere. d) The mean value theorem does not apply to functions with absolute values.

There is at least one mistake.
For example, choice (a) should be False.
$f\left(x\right)$ is continuous for all $x$.
There is at least one mistake.
For example, choice (b) should be True.
$f\left(x\right)$ is not differentiable at $x=1$ and the mean value theorem only applies to functions which are differentiable on some closed interval.
There is at least one mistake.
For example, choice (c) should be False.
$f\left(x\right)$ is continuous for all $x$.
There is at least one mistake.
For example, choice (d) should be False.
As long as the function is differentiable on some closed interval, the Mean Value Theorem applies, even if absolute values appear in the formula for the function. In particular, the mean value theorem does apply to the function $f\left(x\right)=|x-1|$ on, say, the closed interval $\left[2,3\right]$, where it is differentiable.
Correct!
1. False $f\left(x\right)$ is continuous for all $x$.
2. True $f\left(x\right)$ is not differentiable at $x=1$ and the mean value theorem only applies to functions which are differentiable on some closed interval.
3. False $f\left(x\right)$ is continuous for all $x$.
4. False As long as the function is differentiable on some closed interval, the Mean Value Theorem applies, even if absolute values appear in the formula for the function. In particular, the mean value theorem does apply to the function $f\left(x\right)=|x-1|$ on, say, the closed interval $\left[2,3\right]$, where it is differentiable.
Let $f:ℝ\to ℝ$ be a function that is differentiable everywhere. If $f\left(1\right)=-3$ and ${f}^{\prime }\left(x\right)>5$, for all $x$, what is the best lower estimate for $f\left(6\right)$ given by the mean value theorem? Exactly one option must be correct)
 a) $f\left(6\right)>22$ b) $f\left(6\right)>25$ c) $f\left(6\right)<25$ d) $f\left(6\right)>8$ e) $f\left(6\right)<22$

Choice (a) is correct!
By mean value theorem
$\frac{f\left(6\right)-f\left(1\right)}{5}={f}^{\prime }\left(c\right)>5$
so $f\left(6\right)>22$.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Choice (e) is incorrect