The mean value theorem and Rolle's theorem

Worked solutions

Problem 1

1. $f$ is a polynomial, so it is continuous and differentiable everywhereBoth hypotheses hold on any interval.
2. $f(-2) = -8 + 8 = 0$ and $f(2) = 8 - 8 = 0$The endpoint values agree.
3. $f'(x) = 3x^2 - 4$Power rule.
4. $3c^2 - 4 = 0$ gives $c = \pm 2/\sqrt3$Solve $c^2 = 4/3$.
5. $2/\sqrt3 \approx 1.1547$, inside $(-2, 2)$Both signs are in the interval, so both count.
6. $c = \pm 2/\sqrt3 \approx \pm 1.1547$Rolle promised at least one; this function has two, one at each turning point between its zeros at $-2$, $0$ and $2$.

Problem 2

1. Secant slope $= \dfrac{27 - 0}{3 - 0} = 9$$f(3) = 27$ and $f(0) = 0$.
2. $f'(c) = 3c^2 = 9$, so $c = \pm\sqrt3$Power rule, then solve $c^2 = 3$.
3. $-\sqrt3$ is not in $(0, 3)$The theorem only speaks about points strictly inside the interval.
4. $c = \sqrt3 \approx 1.732$At this one point the tangent to $y = x^3$ is parallel to the secant from $(0, 0)$ to $(3, 27)$.

Problem 3

1. $f$ is continuous on $[1, 9]$ and differentiable on $(1, 9)$$\sqrt x$ is differentiable for $x > 0$; the trouble at $0$ is outside this interval.
2. Secant slope $= \dfrac{3 - 1}{9 - 1} = \dfrac14$$\sqrt9 = 3$ and $\sqrt1 = 1$.
3. $f'(c) = \dfrac1{2\sqrt c} = \dfrac14$Power rule with exponent $\tfrac12$.
4. $\sqrt c = 2$, so $c = 4$Cross-multiply, then square.
5. $c = 4$, inside $(1, 9)$The point is not the midpoint $5$: $\sqrt x$ is concave, so its slope falls, and the average slope is reached left of center.

Problem 4

1. For $f$: slope $= \dfrac{2 - 1}{2 - (-1)} = \dfrac13$$|2| = 2$ and $|-1| = 1$.
2. $f'(x) = -1$ for $x < 0$ and $1$ for $x > 0$; there is no derivative at $0$The two one-sided slopes at the corner differ.
3. So $f'$ never equals $\tfrac13$Its only values are $\pm1$.
4. For $g$: slope $= \dfrac{1 - (-1)}{1 - (-1)} = 1$$g(1) = 1$ and $g(-1) = -1$.
5. $g'(x) = -1/x^2 < 0$ wherever it is defined, so it never equals $1$Power rule; a negative number is not $1$.
6. $f$: slope $\tfrac13$, but $f' = \pm1$ because $f$ is not differentiable at $0$; $g$: slope $1$, but $g' < 0$ because $g$ is not continuous at $0$ (not even defined there)Each fails one hypothesis at one point, and that is enough for the conclusion to fail. The theorem is not wrong; it does not apply.

Problem 5

1. $\sqrt{101} - \sqrt{100} = \dfrac{1}{2\sqrt c}\,(101 - 100) = \dfrac1{2\sqrt c}$ for some $c$ in $(100, 101)$MVT; $f'(x) = 1/(2\sqrt x)$.
2. $c > 100$, so $\dfrac1{2\sqrt c} < \dfrac1{2 \cdot 10} = 0.05$A larger $c$ gives a smaller fraction.
3. So $\sqrt{101} < 10.05$The upper bound, from the left end of the interval.
4. $\sqrt c < \sqrt{101} < 10.05$, so $\dfrac1{2\sqrt c} > \dfrac1{20.1}$Use the upper bound just found on the right end.
5. $10 + \tfrac1{20.1} \approx 10.049751 < \sqrt{101} < 10.05$The true value is $10.049876$. The upper bound uses $c > 100$ and the lower bound $c < 101$; which end gives which bound depends on the derivative falling.

Problem 6

1. $\sin a - \sin b = \cos c\,(a - b)$ for some $c$ between $a$ and $b$MVT on $\sin$, which is differentiable everywhere; for $a = b$ both sides are $0$.
2. $|\cos c| \le 1$, so $|\sin a - \sin b| \le |a - b|$$\sin$ is $1$-Lipschitz.
3. For $x > 0$: $\sin x - \sin 0 = x\cos c$ with $0 < c < x$MVT on $[0, x]$.
4. If $x \le 2\pi$ then $c$ is in $(0, 2\pi)$, so $\cos c < 1$ and $\sin x < x$; if $x > 2\pi$ then $\sin x \le 1 < x$$\cos c = 1$ only at multiples of $2\pi$.
5. $\sin 1.2 - \sin 1.1 = 0.1\cos c$ with $1.1 < c < 1.2$MVT on $[1.1, 1.2]$.
6. $\cos$ is decreasing on $[0, \pi]$, so $\cos 1.2 < \cos c < \cos 1.1$$\cos' = -\sin < 0$ there.
7. $|\sin a - \sin b| \le |a - b|$ and $\sin x < x$ for $x > 0$; $0.0362 < \sin 1.2 - \sin 1.1 < 0.0454$$0.1\cos 1.2 \approx 0.03624$ and $0.1\cos 1.1 \approx 0.04536$. The true difference is $0.04083$. The crude bound $0.1$ from step 2 is true but more than twice too large.

Problem 7

1. Let $f(x) = 2x - \cos x$, continuous and differentiable everywhereA solution of the equation is a zero of $f$.
2. $f(0) = -1 < 0$ and $f(\pi/2) = \pi > 0$$\cos 0 = 1$ and $\cos(\pi/2) = 0$.
3. By the IVT, $f$ has a zero in $(0, \pi/2)$$f$ is continuous and changes sign. This is existence.
4. $f'(x) = 2 + \sin x \ge 1$$\sin x \ge -1$.
5. If $f(r) = f(s) = 0$ with $r < s$, Rolle gives $f'(c) = 0$ for some $c$ in $(r, s)$, which step 4 rules outThis is uniqueness: two zeros force a flat point between them.
6. exactly one solution, in $(0, \pi/2)$; it is $x \approx 0.4502$Newton's method from $0.4$ gives $0.450184$. The IVT says there is at least one root and Rolle says there is at most one; neither theorem alone says exactly one.

Problem 8

1. Let $s(t)$ be the distance traveled in km at time $t$ in hours after 10:00, with $s(0) = 0$ and $s(0.1) = 12$Six minutes is $0.1$ hours.
2. Average speed $= \dfrac{12 - 0}{0.1 - 0} = 120$ km/hThe secant slope of $s$ on $[0, 0.1]$.
3. Assume $s$ is continuous on $[0, 0.1]$ and differentiable on $(0, 0.1)$A car's position does not jump and it has a speed at every instant.
4. By the MVT, $s'(c) = 120$ for some $c$ in $(0, 0.1)$The speedometer reads the derivative.
5. At some instant between 10:00 and 10:06 the car was going exactly $120$ km/h, above the $100$ km/h limitThe theorem gives the instant's existence, not its time. Averaging $120$ km/h without ever exceeding $100$ km/h would need a position that is not differentiable, a car that teleports.

Problem 9

1. Let $h = f - g$, so $h' = 0$ on the intervalDifference rule.
2. For any $x < y$ in the interval, $h(y) - h(x) = h'(c)(y - x) = 0$MVT on $[x, y]$, which lies inside the interval because an interval has no gaps.
3. So $h$ takes the same value everywhere: $f = g + C$Any two points give equal values.
4. Let $F(x) = \arctan x + \arctan(1/x)$. Then $F'(x) = \dfrac1{1 + x^2} + \dfrac{-1/x^2}{1 + 1/x^2}$Chain rule on $\arctan(1/x)$, with $(1/x)' = -1/x^2$.
5. $\dfrac{-1/x^2}{1 + 1/x^2} = \dfrac{-1}{x^2 + 1}$, so $F'(x) = 0$ for every $x \ne 0$Multiply top and bottom by $x^2$.
6. $(0, \infty)$ is an interval, so $F$ is constant there, and $F(1) = \tfrac\pi4 + \tfrac\pi4 = \tfrac\pi2$Evaluate at a convenient point.
7. $\arctan x + \arctan(1/x) = \pi/2$ for $x > 0$, and $-\pi/2$ for $x < 0$$(-\infty, 0)$ is another interval with its own constant: $F(-1) = -\tfrac\pi4 - \tfrac\pi4$. The domain of $F$ has a gap at $0$, so the two halves need not agree. The same step is why every antiderivative of $f$ on an interval is one antiderivative plus a constant.

Problem 10

1. $f'(w) = \dfrac{e^w}{1 + e^w} = \sigma(w)$, which lies in $(0, 1)$Chain rule; divide top and bottom by $e^w$.
2. $|f(u) - f(v)| = \sigma(c)\,|u - v| \le |u - v|$MVT, with $0 < \sigma(c) < 1$. So $f$ is $1$-Lipschitz.
3. $f''(w) = \sigma(w)(1 - \sigma(w)) \le \tfrac14$$p(1 - p)$ is largest at $p = \tfrac12$, where it is $\tfrac14$.
4. $|\sigma(u) - \sigma(v)| = \sigma'(c)\,|u - v| \le \tfrac14|u - v|$MVT applied to $f' = \sigma$. So $f'$ is $\tfrac14$-Lipschitz.
5. $f(2) - f(0) = \log(1 + e^2) - \log 2 \approx 2.1269 - 0.6931 = 1.4338 \le 2$The first bound, with $|u - v| = 2$.
6. $\sigma(2) - \sigma(0) \approx 0.8808 - 0.5 = 0.3808 \le 0.5$The second bound: $\tfrac14 \cdot 2 = 0.5$.
7. $\sigma(c) = 1.4338/2 = 0.7169$, so $c = \log\dfrac{0.7169}{1 - 0.7169} \approx 0.929$Invert the sigmoid: $\sigma(c) = p$ gives $c = \log(p/(1 - p))$.
8. $f$ is $1$-Lipschitz and $f' = \sigma$ is $\tfrac14$-Lipschitz; $1.4338 \le 2$, $0.3808 \le 0.5$, and $c \approx 0.929$ in $(0, 2)$The first bound says one example's loss changes by at most as much as its margin does. The second bounds how fast the gradient can change, the constant that step-size rules for gradient descent are stated in terms of.