Corrected exercises: Rolle's theorem and the Mean Value Theorem (MATH 140)

a) Hypothesis (1): $f$ is a polynomial, so it is continuous on $\mathbb{R}$, in particular on $[-2, 2]$. Hypothesis (2): a polynomial is differentiable on $\mathbb{R}$, in particular on $(-2, 2)$. Hypothesis (3): $f(-2) = 16 - 8 = 8$ and $f(2) = 16 - 8 = 8$. The theorem applies. By the power rule, $f'(x) = 4x^3 - 4x = 4x(x - 1)(x + 1)$, which vanishes at $x = -1$, $0$ and $1$, all three in $(-2, 2)$. So $c \in \{-1, 0, 1\}$: the theorem promised AT LEAST one number and there are three, the two low points at height $-1$ and the local high point at the origin that the figure shows. Writing only the hypotheses that happen to be easy, or writing them without a reason (polynomial, hence continuous), loses the method marks even with the right $c$.

b) Sine and cosine are continuous and differentiable on $\mathbb{R}$, so (1) and (2) hold on $\left[0, \frac{\pi}{2}\right]$ and $\left(0, \frac{\pi}{2}\right)$. For (3): $g(0) = 0 + 1 = 1$ and $g\left(\frac{\pi}{2}\right) = 1 + 0 = 1$. Then $g'(x) = \cos x - \sin x = 0$ means $\sin x = \cos x$; on $\left(0, \frac{\pi}{2}\right)$, $\cos x > 0$, so we may divide: $\tan x = 1$, and $x = \frac{\pi}{4}$. The other solutions of $\tan x = 1$, such as $\frac{5\pi}{4}$, lie outside the interval and are not answers. At $c = \frac{\pi}{4}$ the curve reaches $g\left(\frac{\pi}{4}\right) = \frac{\sqrt 2}{2} + \frac{\sqrt 2}{2} = \sqrt 2$, its top on the interval.

c) The fifth root is defined for every real number, so $h(x) = \left(x^{1/5}\right)^4$ is continuous on $[-1, 1]$, and $h(-1) = (-1)^4 = 1 = h(1)$: hypotheses (1) and (3) hold. For $x \ne 0$, the power rule gives $h'(x) = \frac{4}{5}x^{-1/5} = \frac{4}{5x^{1/5}}$, a fraction with numerator $4$, never $0$. At $x = 0$, the difference quotient is $\frac{h(t) - h(0)}{t} = t^{-1/5}$, which tends to $+\infty$ as $t \to 0^+$ and to $-\infty$ as $t \to 0^-$: $h'(0)$ does not exist. Hypothesis (2) fails at ONE point, $0$, inside $(-1, 1)$, and that is enough to lose the conclusion. The graph has a cusp at the origin: a lowest point without a horizontal tangent.

d) No contradiction, because the theorem does not apply. $\tan x = \frac{\sin x}{\cos x}$ is not defined at $x = \frac{\pi}{2}$, which lies in $[0, \pi]$, and $|\tan x| \to \infty$ near it: hypothesis (1) fails (and (2) with it). The equal values at the ends are only one hypothesis out of three. Here $\sec^2 x = \frac{1}{\cos^2 x} \ge 1$ wherever it is defined, so the slope is never $0$, and it cannot be, since the graph is made of two separate branches. The trap is to read hypothesis (3) and stop there.

e) Hypothesis (3): $k(0) = 0 = k(1)$. Hypothesis (2): on $(0, 1)$, $k(x) = x$, so $k$ is differentiable there with $k'(x) = 1$. Hypothesis (1) fails at the right endpoint: $\lim_{x \to 1^-} k(x) = 1 \ne 0 = k(1)$, so $k$ is not continuous at $1$. The conclusion is false, since $k'(x) = 1 \ne 0$ everywhere in $(0, 1)$. The closed bracket in hypothesis (1) is not decoration: continuity AT the endpoints is what forbids the graph from jumping back down to its starting height at the last moment.

a) $f$ is a polynomial, hence continuous on $[0, 2]$ and differentiable on $(0, 2)$. The slope of the chord is $\frac{f(2) - f(0)}{2 - 0} = \frac{6 - 0}{2} = 3$. By the power rule $f'(c) = 3c^2 - 1$, and $3c^2 - 1 = 3$ gives $c^2 = \frac{4}{3}$, so $c = \pm\frac{2}{\sqrt 3} = \pm\frac{2\sqrt 3}{3}$. The theorem speaks about $c$ in $(0, 2)$ only: $-\frac{2\sqrt 3}{3}$ is rejected, and $\frac{2\sqrt 3}{3}$ is kept because it is positive and $\left(\frac{2\sqrt 3}{3}\right)^2 = \frac{4}{3} < 4$, so it is less than $2$. Answering with both values is the most common loss on this question: solving $f'(c) = m$ is algebra, choosing $c$ is the theorem. The solution figure shows the tangent at $c$, parallel to the chord.

b) $e^x$ is continuous and differentiable on $\mathbb{R}$. The chord has slope $\frac{e^1 - e^0}{1 - 0} = e - 1$, and $f'(c) = e^c$, so $e^c = e - 1$ and $c = \ln(e - 1)$. That $e - 1 > 0$ is needed for the logarithm to exist. Now $2 < e$, so $e - 1 > 1$, and obviously $e - 1 < e$. Since $\ln$ is increasing, $\ln 1 < \ln(e - 1) < \ln e$, that is $0 < c < 1$. No decimal is required: the exact answer is $\ln(e - 1)$, and the inequality is proved with $e > 2$ alone.

c) $\ln x$ is continuous and differentiable on $(0, \infty)$, which contains $[1, e]$. The chord has slope $\frac{\ln e - \ln 1}{e - 1} = \frac{1}{e - 1}$, and $f'(c) = \frac{1}{c}$, so $\frac{1}{c} = \frac{1}{e - 1}$ and $c = e - 1$, which lies in $(1, e)$ by the same two inequalities as in b). The two answers are linked: $\ln$ is the inverse of $\exp$, the graph of $\ln$ on $[1, e]$ is the mirror image of the graph of $e^x$ on $[0, 1]$ in the line $y = x$, the chord is reflected into a chord and the tangent into a tangent, so the point of contact $(\ln(e - 1), e - 1)$ becomes $(e - 1, \ln(e - 1))$.

d) The chord has slope $\frac{f(1) - f(-1)}{1 - (-1)} = \frac{1 - (-1)}{2} = 1$, but $f'(x) = -\frac{1}{x^2}$ is negative wherever it exists: no $c$ can give $1$. No contradiction: $\frac{1}{x}$ is not defined at $0$, which lies in $[-1, 1]$, so $f$ is not continuous on $[-1, 1]$ and hypothesis (1) fails. The graph goes down on each branch and jumps from $-\infty$ to $+\infty$ at $0$, and it is the jump that lets the endpoints be in increasing order.

e) $x^{1/3}$ is continuous on $\mathbb{R}$, but at $0$ the difference quotient $\frac{t^{1/3}}{t} = t^{-2/3} \to +\infty$: vertical tangent, no derivative, so hypothesis (2) fails at $0 \in (-1, 1)$. Yet the chord has slope $\frac{1 - (-1)}{2} = 1$ and, for $x \ne 0$, $f'(x) = \frac{1}{3x^{2/3}} = 1$ gives $x^{2/3} = \frac{1}{3}$, so $|x| = \left(\frac{1}{3}\right)^{3/2} = \frac{1}{3\sqrt 3} = \frac{\sqrt 3}{9}$, and $c = \pm\frac{\sqrt 3}{9}$, both in $(-1, 1)$. The conclusion holds although a hypothesis fails. The hypotheses are SUFFICIENT, not necessary: when one fails, the theorem simply guarantees nothing, and the conclusion may hold, as here, or fail, as in d).

a) Take any two numbers $x_1 < x_2$ in $(a, b)$. The segment $[x_1, x_2]$ lies inside $(a, b)$, because $(a, b)$ is an interval; there $f$ is differentiable, hence continuous on $[x_1, x_2]$ and differentiable on $(x_1, x_2)$. The Mean Value Theorem gives $c$ in $(x_1, x_2)$ with $f(x_2) - f(x_1) = f'(c)(x_2 - x_1)$. Since $c$ is in $(a, b)$, $f'(c) = 0$, so $f(x_2) = f(x_1)$. Any two values of $f$ are equal: $f$ is constant. The unknown $c$ is never computed; all we need is that it lands where $f' = 0$. For the second consequence, apply this to $h = f - g$, whose derivative is $f' - g' = 0$.

b) Let $F(x) = \arcsin x + \arccos x$. On $(-1, 1)$, $F'(x) = \frac{1}{\sqrt{1 - x^2}} + \left(-\frac{1}{\sqrt{1 - x^2}}\right) = 0$, and $(-1, 1)$ is an interval, so $F$ is constant there, equal to $F(0) = 0 + \frac{\pi}{2} = \frac{\pi}{2}$. The endpoints need their own line, since the derivatives do not exist at $\pm 1$: $F(1) = \frac{\pi}{2} + 0$ and $F(-1) = -\frac{\pi}{2} + \pi = \frac{\pi}{2}$. The identity holds on all of $[-1, 1]$. The shape of the proof is the one to learn: derivative zero, constant on the interval, the constant read at one convenient point.

c) Let $g(x) = f(x) - x^3$. By the difference and power rules, $g'(x) = 3x^2 - 3x^2 = 0$ for every $x$ in $\mathbb{R}$, which is an interval, so $g$ is a constant $C$. At $x = 2$: $C = g(2) = f(2) - 8 = 5 - 8 = -3$. Hence $f(x) = x^3 - 3$ for every $x$, and no other function fits. This is the uniqueness behind every initial value problem: two functions with the same derivative on an interval differ by a constant, and one value fixes it.

d) $f''(x) = (f')'(x) = 0$ for every real $x$, so by a) applied to $f'$ on $\mathbb{R}$, $f'$ is a constant, call it $m$. Then $k(x) = f(x) - mx$ has $k'(x) = m - m = 0$ on $\mathbb{R}$, so $k$ is a constant $b$, and $f(x) = mx + b$. The consequence is used twice, once per derivative: a function whose second derivative vanishes has a graph that is a straight line, and the proof says why, instead of saying that it is obvious.

e) Let $u = \frac{1 + x}{1 - x}$. The quotient rule gives $u' = \frac{(1)(1 - x) - (1 + x)(-1)}{(1 - x)^2} = \frac{2}{(1 - x)^2}$, and $1 + u^2 = \frac{(1 - x)^2 + (1 + x)^2}{(1 - x)^2} = \frac{2 + 2x^2}{(1 - x)^2}$. By the chain rule with inner function $u$, $\frac{d}{dx}\arctan u = \frac{u'}{1 + u^2} = \frac{2}{2 + 2x^2} = \frac{1}{1 + x^2}$, so $D'(x) = \frac{1}{1 + x^2} - \frac{1}{1 + x^2} = 0$ for $x \ne 1$. At $x = \sqrt 3$: $\frac{1 + \sqrt 3}{1 - \sqrt 3} = \frac{(1 + \sqrt 3)^2}{1 - 3} = -(2 + \sqrt 3)$, and $\tan\frac{5\pi}{12} = \tan\left(\frac{\pi}{4} + \frac{\pi}{6}\right) = \frac{1 + \frac{1}{\sqrt 3}}{1 - \frac{1}{\sqrt 3}} = \frac{\sqrt 3 + 1}{\sqrt 3 - 1} = 2 + \sqrt 3$, so $\arctan(-(2 + \sqrt 3)) = -\frac{5\pi}{12}$. Hence $D(\sqrt 3) = -\frac{5\pi}{12} - \frac{\pi}{3} = -\frac{3\pi}{4} \ne \frac{\pi}{4}$. The student's error: the domain $(-\infty, 1) \cup (1, \infty)$ is NOT an interval, and the Mean Value Theorem cannot be applied on $[0, \sqrt 3]$, which contains the point $1$ where $D$ is not even defined. $D$ is constant on EACH piece: $\frac{\pi}{4}$ on the left, $-\frac{3\pi}{4}$ on the right, as the solution figure shows.

a) If $a = b$ both sides are $0$. Otherwise, by symmetry of the inequality we may assume $a < b$. Sine is continuous and differentiable on $\mathbb{R}$, so the Mean Value Theorem on $[a, b]$ gives $c$ in $(a, b)$ with $\sin b - \sin a = \cos c\,(b - a)$. Whatever $c$ is, $|\cos c| \le 1$, hence $|\sin b - \sin a| = |\cos c|\,|b - a| \le |b - a|$. With $b = 0$: $|\sin a| \le |a|$ for every $a$. The student who tries to find $c$ is stuck: it depends on $a$ and $b$ and has no formula. The whole method is to bound $f'$ on the interval and let $c$ be anywhere in it.

b) Fix $x > 0$. $\arctan$ is continuous and differentiable on $\mathbb{R}$, with derivative $\frac{1}{1 + t^2}$. The Mean Value Theorem on $[0, x]$ gives $c$ in $(0, x)$ with $\arctan x - \arctan 0 = \frac{x}{1 + c^2}$. From $0 < c < x$: $1 < 1 + c^2 < 1 + x^2$, so $\frac{1}{1 + x^2} < \frac{1}{1 + c^2} < 1$, and multiplying by $x > 0$ keeps the directions: $\frac{x}{1 + x^2} < \arctan x < x$. The inequalities are strict because $c$ is STRICTLY between $0$ and $x$. The figure shows the three curves in that order for every $x > 0$, touching only at the origin.

c) With $x = \frac{1}{\sqrt 3}$: $\arctan\frac{1}{\sqrt 3} = \frac{\pi}{6}$, and $\frac{x}{1 + x^2} = \frac{1/\sqrt 3}{4/3} = \frac{3}{4\sqrt 3} = \frac{\sqrt 3}{4}$. So $\frac{\sqrt 3}{4} < \frac{\pi}{6} < \frac{1}{\sqrt 3} = \frac{\sqrt 3}{3}$, and multiplying by $6$: $\frac{3\sqrt 3}{2} < \pi < 2\sqrt 3$. In decimals, about $2.60 < \pi < 3.46$: a crude bracket, but a proved one, obtained by hand from a derivative bound. The same method on a smaller angle gives a sharper bracket, since $c$ is then trapped in a shorter interval.

d) $f$ is differentiable on $\mathbb{R}$, hence continuous on $[2, 7]$ and differentiable on $(2, 7)$. By the Mean Value Theorem, $f(7) - f(2) = f'(c)(7 - 2) = 5f'(c)$ for some $c$ in $(2, 7)$, and $f'(c) \le 5$, so $f(7) - f(2) \le 25$ and $f(7) \le -3 + 25 = 22$. The bound is reached: $f(x) = 5x - 13$ has $f(2) = -3$, $f'(x) = 5 \le 5$ and $f(7) = 22$. So the largest possible value is $22$. A bound is only the answer to how large can it be when it is shown to be attained: that second line earns its own mark.

e) Fix $x > 0$ and let $f(t) = \sqrt{1 + t}$, continuous on $[0, x]$ and differentiable on $(0, x)$ since $1 + t > 0$ there, with $f'(t) = \frac{1}{2\sqrt{1 + t}}$ by the chain rule. The Mean Value Theorem gives $c$ in $(0, x)$ with $\sqrt{1 + x} - 1 = \frac{x}{2\sqrt{1 + c}}$. Since $c > 0$, $\sqrt{1 + c} > 1$, so $\frac{x}{2\sqrt{1 + c}} < \frac{x}{2}$, and $\sqrt{1 + x} < 1 + \frac{x}{2}$. Check at $x = 3$: $\sqrt 4 = 2 < 2.5$. The same four lines prove every inequality of the form $f(x) - f(0)$ against $x$ times a bound: choose $f$, name the interval, apply the theorem, bound $f'(c)$ using only where $c$ lies.