The Mean Value Theorem for MATH 139

Step 1: Verify the conditions

$f(x) = x^2$ is a polynomial, so it's continuous on $[1, 3]$ and differentiable on $(1, 3)$. ✓

Step 2: Calculate the average rate of change

$$\frac{f(3) - f(1)}{3 - 1} = \frac{9 - 1}{2} = \frac{8}{2} = 4$$

Step 3: Find $c$ where $f'(c) = 4$

$$f'(x) = 2x$$ $$2c = 4$$ $$c = 2$$

Step 4: Verify $c$ is in the interval

$c = 2$ is in $(1, 3)$ ✓

$$\boxed{c = 2}$$

Step 1: Assume (for contradiction) there are two solutions

Suppose $f(x) = x^3 + 3x + 1 = 0$ has two solutions $a$ and $b$ with $a < b$.

Then $f(a) = 0$ and $f(b) = 0$.

Step 2: Apply the Mean Value Theorem

Since $f$ is a polynomial, it's continuous on $[a, b]$ and differentiable on $(a, b)$.

By MVT, there exists $c$ in $(a, b)$ such that: $$f'(c) = \frac{f(b) - f(a)}{b - a} = \frac{0 - 0}{b - a} = 0$$

Step 3: Check if $f'(x) = 0$ is possible

$$f'(x) = 3x^2 + 3 = 3(x^2 + 1)$$

Since $x^2 + 1 \geq 1 > 0$ for all $x$, we have $f'(x) > 0$ always.

Step 4: Contradiction!

We needed $f'(c) = 0$, but $f'(x) > 0$ everywhere. This is impossible!

$$\boxed{\text{Therefore, } x^3 + 3x + 1 = 0 \text{ has at most one solution}}$$

Step 1: We already know "at most one" from Example 2

Step 2: Show "at least one" using the Intermediate Value Theorem

$f(x) = x^3 + 3x + 1$

$f(-1) = -1 - 3 + 1 = -3 < 0$

$f(0) = 0 + 0 + 1 = 1 > 0$

Since $f$ is continuous and changes sign, by IVT there's at least one root in $(-1, 0)$.

$$\boxed{\text{Exactly one solution (at most one + at least one)}}$$

Step 1: Apply MVT to $[0, 3]$

By MVT, there exists $c$ in $(0, 3)$ such that: $$f'(c) = \frac{f(3) - f(0)}{3 - 0} = \frac{f(3) - 5}{3}$$

Step 2: Use the bound on $f'$

Since $f'(c) \leq 2$: $$\frac{f(3) - 5}{3} \leq 2$$ $$f(3) - 5 \leq 6$$ $$f(3) \leq 11$$

$$\boxed{f(3) \leq 11}$$

Interpretation: If your "speed" is at most 2, and you start at height 5, then after 3 units of time, you can be at most at height 11.

Step 1: Set up the problem

Let $s(t)$ = position at time $t$ (in hours after 2:00 PM)

$s(0) = 0$ miles (entered toll road)

$s(1) = 100$ miles (exited after 1 hour)

Step 2: Apply MVT

By MVT, there exists some time $c$ in $(0, 1)$ where: $$s'(c) = \frac{s(1) - s(0)}{1 - 0} = \frac{100 - 0}{1} = 100 \text{ mph}$$

Step 3: Conclusion

At some moment, the instantaneous speed was exactly 100 mph.

Since $100 > 70$, the driver definitely exceeded the speed limit.

$$\boxed{\text{Yes, the driver can be proven to have sped}}$$

Step 1: Let $f(x) = x - \sin x$ and show $f(x) > 0$ for $x > 0$

$f(0) = 0 - 0 = 0$

Step 2: Apply MVT to $[0, x]$ for any $x > 0$

By MVT, there exists $c$ in $(0, x)$ such that: $$f'(c) = \frac{f(x) - f(0)}{x - 0} = \frac{f(x)}{x}$$

Step 3: Analyze $f'$

$$f'(t) = 1 - \cos t$$

Since $\cos t \leq 1$ for all $t$, we have $f'(t) \geq 0$ for all $t$.

For $c \in (0, x)$, we have $\cos c < 1$ (since $c \neq 0$), so $f'(c) > 0$.

Step 4: Conclude

$$\frac{f(x)}{x} = f'(c) > 0$$

Since $x > 0$, we get $f(x) > 0$, meaning: $$x - \sin x > 0$$ $$\boxed{\sin x < x \text{ for all } x > 0}$$