Important theorems

$f(x)$ is a polynomial so it’s continuous and differentiable everywhere.

The average rate of change over $[0,3]$ is:

$$\frac{f(3)-f(0)}{3-0}$$

$$=\frac{6-0}{3}$$

$$=2$$

The derivative of $f(x)$ is

$$f^{\prime }(x)=3x^2-6x+2$$

At some point in the interval, $f^{\prime }(x)=2$.

$$3x^2-6x+2=2$$

$$3x^2-6x=0$$

$$3x(x-2)=0$$

$$x=0,2$$

Because $x=0$ is an endpoint and not included in the open interval $(0,3)$, it must be eliminated. So $\boxed{c=2}$ is the only value that satisfies the MVT.

$f(x)$ is the top half of the unit circle. It is continuous on the interval $[-1,0]$ even though at the endpoint $x=-1$, it doesn’t have a two-sided limit.

This isn’t a problem because continuity on a closed interval $[a,b]$ only requires the function to be continuous from the right at the left endpoint i.e. $\underset{x\to a^{+}}{\lim }f(x)=f(a)$, and continuous from the left and the right endpoint i.e. $\underset{x\to b^{-}}{\lim }f(x)=f(b)$.

While the derivative does not exist at the endpoint $x=-1$, differentiability over the open interval is sufficient for the MVT to apply. Since $f(x)$ is differentiable over $(-1,0)$, both conditions of the MVT have been met.

The average rate of change over $[-1,0]$ is

$$\frac{f(0)-f(-1)}{0-(-1)}$$

$$=\frac{1-0}{1}$$

$$=1$$

The derivative of $f(x)$ is

$$f^{\prime }(x)=-\frac{x}{\sqrt{1-x^2}}$$

At some point in the interval, $f^{\prime }(x)=1$.

$$-\frac{x}{\sqrt{1-x^2}}=1$$

$$-x=\sqrt{1-x^2}$$

$$x^2=1-x^2$$

$$2x^2=1$$

$$x=\pm \frac{\sqrt{2}}{2}$$

During the equation-solving process, both sides were squared, which introduced an extraneous solution. Plugging in both values yields $\boxed{x=-\frac{\sqrt{2}}{2}}$ as the only solution, or the value of $c$, that satisfies the MVT.

The average speed is the distance divided by the change in time, or

$$\frac{100\text{ miles}}{2\text{ hours}}=50\text{ mph}$$

Assuming the car’s position is a continuous and differentiable function of time (reasonable for real-life driving), then the MVT guarantees that at some point, the car’s instantaneous speed was 50 mph, which exceeds the posted speed limit.

$f(x)$ is a rational function and discontinuous at $x=3$, which is in the given interval. So Rolle’s theorem can’t be applied. Even though a horizontal tangent may exist within the interval, the theorem does not guarantee this.

$f(x)$ is continuous but not differentiable at $x=0$, which is in the given interval. So Rolle’s theorem does not apply and indeed, there’s no point on the graph where the tangent line is horizontal.

1. Verify continuity

$f(x)$ is a polynomial and continuous.

2. Find critical points

$$f^{\prime }(x)=2x-4$$

Solving for $f^{\prime }(x)=0$,

$$2x-4=0$$

$$x=2$$

3. Evaluate function values

At the critical point $x=2$,

$$\begin{gathered} f(2)=2^2-4(2)+3 \\ =-1 \end{gathered}$$

At the two endpoints,

$$\begin{gathered} f(0)=0^2-4(0)+3 \\ =3 \end{gathered}$$

$$\begin{gathered} f(5)=5^2-4(5)+3 \\ =8 \end{gathered}$$

4. Compare function values

The absolute maximum is $8$ which occurs at the point $x=5$.

The absolute minimum is $-1$ at $x=2$.

1. Verify continuity

The rational function has an asymptote at $x=-1$ only but is continuous on the given interval.

2. Find critical points

$$f^{\prime }(x)=\frac{(x+1)(1)-(x)(1)}{(x+1{)}^2}$$

$$=\frac{1}{(x+1{)}^2}$$

Since $f^{\prime }(x)>0$ everywhere, there are no critical points. Not every function has to have critical points!

3. Evaluate function values

Only the values at the endpoints need to be evaluated.

$$f(0)=\frac{0}{0+1}$$

$$=0$$

$$f(3)=\frac{3}{3+1}$$

$$=\frac{3}{4}$$

4. Compare function values.

The absolute minimum is $0$ at $x=0$.

The absolute maximum is $\frac{3}{4}$ at $x=3$.