Find the equation of the tangent line to
$y=\sqrt{{e}^{5x}+3}$ at
$x=0$. Exactly one option
must be correct)

*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)

*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)

*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)

*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)

For example, choice (a) should be False.

For example, choice (b) should be True.

For example, choice (c) should be False.

For example, choice (d) should be False.

For example, choice (e) should be True.

*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!*

*False**True**False**False**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)
*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 \pm \infty}f\left(x\right)=\pm \infty $ the function
$f\left(x\right)$ must have at least
one root. If $f\left(x\right)$ has two
roots, say $a<b$, 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)

For example, choice (a) should be False.

For example, choice (b) should be False.

For example, choice (c) should be True.

*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!*

*False**False**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)
*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

For example, choice (a) should be False.

For example, choice (b) should be True.

For example, choice (c) should be False.

For example, choice (d) should be False.

$${f}^{\prime}\left(c\right)=\frac{f\left(3\right)-f\left(0\right)}{3}.$$

This does not contradict the mean value theorem because (tick each correct
answer):
(Zero or more options can be correct)
*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!*

*False*$f\left(x\right)$ is continuous for all $x$.*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.*False*$f\left(x\right)$ is continuous for all $x$.*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 $f:\mathbb{R}\to \mathbb{R}$
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)

*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*