Since the function $f(x) = 9- (x-3)^2$ is a polynomial, it is continuous on the interval $[0, 6]$ and differentiable on the interval $(0,6)$. Furthermore, as the question states, $f(0) = f(6)$, and so Rolle’s Theorem applies. We’re looking for a value of $c$ such that $f'(c) = 0$:

\begin{align*} f'(x) &= \frac{d}{dx}\left[9- (x-3)^2 \right] \\ \\ &= -2(x-3) \end{align*} We want to find $c$ such that $f'(c) = 0$: \begin{align*} f'(c) &= -2(c-3) = 0 \\ \\ c &= 3 \quad \cmark \end{align*}

The point $c = 3$ has a horizontal tangent line, satisfying Rolle’s Theorem since $0 < c < 6$.

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Since the function $f(x) = x^2$ is a polynomial, it is continuous on the interval $[1, 4]$ and differentiable on the interval $(1,4)$, and so the Mean-Value Theorem applies. We’re looking for a value of $c$ such that \begin{align*} f'(c) &= \frac{f(4) – f(1)}{4-1} \\ \\ &= \frac{16 – 1}{3} \\ \\ &= \frac{15}{3} = 5 \end{align*}

Now we can compute $f'(x)$ starting with $f(x) = x^2$: \begin{align*} f'(x) &= \frac{d}{dx}x^2 \\ \\ &= 2x \end{align*} We want to find $c$ such that $f'(c) = 5$: \begin{align*} f'(c) &= 2c = 5 \\ \\ c &= 2.5 \quad \cmark \end{align*}

The tangent line at the point $c = 2.5$ is parallel to the secant line that connects the endpoints of the interval.

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We’ll follow the hint and use the MVT. Note that $f'(x) \le 3$ for all $x$, so we know $f$ is differentiable (and continuous) for all $x$ and so the MVT applies.

The problem gives us the value of $f$ at $x = 1$ and asks for a conclusion about the value at $x = 5$, so let’s apply the MVT to the interval $[1,5]$:

\begin{align*} f(b) – f(a) &= f'(c)(b-a) \\ \\ f(5) – f(1) &= f'(c)(5-1) \\ \\ f(5) – 2 &= f'(c)(4) \\ \\ f(5) &= 4f'(c) +2 \end{align*} We’re told that $f'(x) \le 3$ for all $x$, so $4f'(c) \le 12$. Hence \begin{align*} \phantom{f(b) – f(a)} & \\ f(5) &\le 12 + 2 \\ \\ &\le 14 \end{align*} The largest possible value of $f(5)$ is 14. $\quad \cmark$

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This question might seem silly: “Prove that if a function has positive slope, then it is increasing.” But mathematics is partly about using theorems you’ve already proven to prove other things, no matter how obvious they may seem, so let’s just do as the question asks.

First, note that since f(x) is differentiable for all x, it must be continuous for all x, and so the Mean Value Theorem (MVT) applies. The problems says to use the MVT, so let’s start there, and consider an interval $(a,b)$. The MVT tells us that there exists a c, $a < c < b$, such that $$f(b) - f(a) = f'(c) (b-a)$$ Since $f'(x) > 0$ for all $x$, we have $f'(c) > 0$. Furthermore, since $b > a$, $b-a > 0$. Hence $$f(b) – f(a) > 0$$ Hence $$f(b) > f(a)$$ That is, $f(x)$ is increasing. $\quad \cmark$

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The problem says to use the Mean Value Theorem, and we’re trying to show something about $\sin(x)$, so let’s apply the MVT to $f(x) = \sin(x)$ on the interval $(a,b)$ and see what happens. Note that $f(x) = \sin(x)$ is continuous and differentiable for all real values of $x$, so the MVT applies.

\begin{align*} \frac{f(b) – f(a)}{b-a} &= f'(c) \\[8px] \frac{\sin(b) – \sin(a)}{b-a} &= f'(c)\\[8px] \end{align*} Now $f'(x) = \cos(x)$, so: \begin{align*} \frac{\sin(b) – \sin(a)}{b-a} &= \cos(c) \end{align*} Furthermore, $|\cos(c)| \le 1$. Hence \begin{align*} \left|\frac{\sin(b) – \sin(a)}{b-a} \right| &= |\cos(c)| \le 1 \\[8px] \left|\frac{\sin(b) – \sin(a)}{b-a} \right| &\le 1 \\[8px] |\sin(b) – \sin(a)| &\le |b-a| \quad \cmark \end{align*}

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The problem says to use the Mean Value Theorem, and we’re trying to show something about $\cos(2x)$, so let’s apply the MVT to $f(x) = \cos(2x)$ on the interval $(a,b)$ and see what happens. Note that $f(x) = \cos(2x)$ is continuous and differentiable for all real values of $x$, so the MVT applies.

\begin{align*} \frac{f(b) – f(a)}{b-a} &= f'(c) \\[8px] \frac{\cos(2b) – \cos(2a)}{b-a} &= f'(c) \end{align*} Now $f'(x) = -2\sin(2x)$, so: \begin{align*} \frac{\cos(2b) – \cos(2a)}{b-a} &= -2\sin(2c) \end{align*} Furthermore, $|\sin(2c)| \le 1$. Hence $|-2\sin(2c)| \le 2$ and so \begin{align*} \left|\frac{\cos(2b) – \cos(2a)}{b-a} \right| &= |-2\sin (2c)| \le 2 \\[8px] \left|\frac{\cos(2b) – \cos(2a)}{b-a} \right| &\le 2 \\[8px] |\cos(2b) – \cos(2a)| &\le 2|b-a| \quad \cmark \end{align*}

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The problem says to use the Mean Value Theorem, and we’re trying to show something about $e^x$, so let’s apply the MVT to $f(x) = e^x$ on the interval $(a,b)$ and see what happens. Note that $e^x$ is continuous and differentiable for all real $x$, so the MVT applies.

\begin{align*} \frac{f(b) – f(a)}{b-a} &= f'(c) \\ \\ \frac{e^b – e^a}{b-a} &= f'(c) \\ \\ e^b – e^a &= f'(c) (b-a) \end{align*} Hmmm. We’re a little stuck there. But looking back at the question, we see that we’re after an inequality that involves $e^x$, so let’s set $b = x$: \begin{align*} \phantom{ \frac{f(b) – f(a)}{b-a}} & \\ e^x – e^a &= f'(c) (x -a) \end{align*} That’s looking more promising, but we need to get rid of the $a$. For lack of other ideas, let’s try $a = 0$ to make it go away. And note that $e^0 = 1$: \begin{align*} \phantom{ \frac{f(b) – f(a)}{b-a}} & \\ e^x – 1 &= f'(c) (x – 0) \\ \\ &= f'(c)x \\ \\ e^x &= f'(c)x + 1 \end{align*} Oh, that’s now close! Let’s think about $f'(c)$: we know $f'(x) = e^x$, so $f'(c) = e^c$: \begin{align*} \phantom{ \frac{f(b) – f(a)}{b-a}} & \\ e^x &= e^c x + 1 \end{align*} But we’re only interested in $x>0$. And for $c > 0$, we know $e^c > 1$. Hence $e^c x > x$, and so \begin{align*} \phantom{ \frac{f(b) – f(a)}{b-a}} & \\ e^x &> x + 1 \quad \cmark \end{align*}

Whew, done! This “proof problem” is a good example of a question where we can’t immediately see how it’s going to turn out when we begin. Instead, you just have to dive in and start, and then try different things to keep moving forward, with the hope you’ll land in the right place at the end. In this case, we had to just invoke that $a = 0$ because we didn’t want an $a$ in our final expression, but we didn’t know that until we hit a certain point along the way. The crucial take away: Start, and then keep going. If one path doesn’t work out, try another!

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There are two parts to this proof:
(I) Show that the function has at least one root. We’ll use the Intermediate Value Theorem for this part and show there exists some value $c_1$ $(-\infty < c_1 < \infty)$ for which $f(c_1) = 0$.
(II) We’ll then use Rolle’s Theorem to show, by contradiction, that $f(x)$ does not have two or more roots.
Those two facts together mean that $f(x)$ has one and only one root.

(I) Let’s first show that the function has one real root.
First note that $\displaystyle{\lim_{x \to -\infty} f(x) = -\infty}$, while $\displaystyle{\lim_{x \to \infty} f(x) = \infty}$. Since $f$ is a continuous function, by the Intermediate Value Theorem there must be some value $c_1$ $(-\infty < c_1 < \infty)$ for which $f(c_1) = 0$. That is, $f(x)$ has at least one root.

(II) A proof by contradiction has three key steps:
1. Assume the opposite of what you want to prove.
2. Use that assumption to develop a new condition.
3. Show that the new condition contradicts your original assumption. Therefore your original assumption must be wrong, and so you’ve proved what you were originally after.

We’ll proceed using those steps:
1. Let’s assume that $f(x)$ has two real roots. We’ll call them $a$ and $b$, such that $f(a) = f(b) = 0$.

2. Rolle’s Theorem applies to $f(x)$, since $f$ is a polynomial and is therefore continuous and differentiable everywhere. According to the Theorem, then, there must be a value $c_2$ on the interval $(a,b)$ such that: $$f'(c_2) = 0$$
3. Let’s compute $f'(x)$: \begin{align*} f'(x) &= \frac{d}{dx}\left[ x^5 + 6x^3 + x – 4 \right] \\[8px] &= 5x^4 + 18x^2 + 1 \end{align*} \begin{align*} f'(x) &\ge 0 + 0 + 1 \\[8px] &\ge 1 &&\text{for all $x$} \end{align*} That is, $f'(x)$ can never be 0. But in Step 2 above we saw that if there are two roots such that $f(a) = f(b) = 0$, then there must be some value $c_2$ such that $f'(c_2) = 0$. So we have on the one hand $f'(x)$ can never be 0, while on the other hand $f'(c_2) = 0$ for some value of $c_2$. This is a contradiction, and so our original assumption in Step 1 must have been incorrect: $f$ cannot have two real roots. (By extension, $f$ cannot have three or more roots either.)

Hence we have shown that (I) $f$ has at least one real root, and (II) it does not have two or more real roots. Therefore $f$ has exactly one real root. $\quad \cmark$

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We’ll take the same approach we did in the preceding problem:
(I) show that the function has at least one root, and then
(II) use Rolle’s Theorem and proof by contradiction to show that it does not have two or more roots.

(I) To show that $g(x)$ has at least one root, let’s again use the Intermediate Value Theorem (IVT). To keep things simple, let’s try some values for $x$ for which $\cos x = 0$: $x = -\pi/2$ and $\pi/2$.

\begin{align*} g(-\pi/2) &= 6(-\dfrac{\pi}{2}) – \cos(\dfrac{\pi}{2}) = -3\pi \text{, which is } < 0 \\[8px] g(\pi/2) &= 6(\dfrac{\pi}{2}) - \cos(\dfrac{\pi}{2}) = 3\pi \text{, which is } > 0 \end{align*}

Since $g(x)$ is a continuous function, we then know from the IVT that there exists $c_1$, $-\dfrac{\pi}{2} < c_1 < \dfrac{\pi}{2}$, such that $g(c_1) = 0$. Thus $g(x)$ has at least one real root.

(II) We again use proof by contradiction, following the same three steps as we did in part (a):
1. Assume the opposite of what you want to prove.
2. Use that assumption to develop a new condition.
3. Show that the new condition contradicts your original assumption. Therefore your original assumption must be wrong, and so you’ve proved what you were originally after.

We’ll proceed using those steps:
1. Let’s assume that $g(x)$ has two real roots. We’ll call them $a$ and $b$, such that $g(a) = g(b) = 0$.

2. Rolle’s Theorem applies to $g(x)$, since $g$ is the sum of a polynomial and the cosine function and is therefore continuous and differentiable everywhere. According to the Theorem, then, there must be a value $c_2$ on the interval $(a,b)$ such that: $$g'(c_2) = 0$$
3. Let’s compute $g'(x)$: \begin{align*} g'(x) &= \frac{d}{dx}[ 6x – \cos x] \\[8px] &= 6 + \sin x \\[8px] &\ge 5 \end{align*} since the most-negative $\sin x$ can be is $-1.$ Hence $g'(x)$ can never be zero. But in Step 2 above we saw that if there are two roots such that $g(a) = g(b) = 0$, then there must be some value $c_2$ such that $g'(c_2) = 0$. So we have on the one hand $g'(x)$ can never be 0, while on the other hand $g'(c_2) = 0$ for some value of $c_2$. This is a contradiction, and so our original assumption in Step 1 must have been incorrect: $g$ cannot have two real roots. (By extension, $g$ cannot have three or more roots either.)

Hence we have shown that (I) $g$ has at least one real root, and (II) it does not have two or more real roots. Therefore $g$ has exactly one real root. $\quad \cmark$

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