Corrected exercises: Rolle's theorem and the mean value theorem (201-SN2-RE)

$f(1)=4$, $f(5)=40$, so $m=\frac{36}{4}=9$. $f'(c)=2c+3=9$ gives $c=3$, the midpoint of $[1, 5]$: for every quadratic function, the $c$ of the mean value theorem is the midpoint of the interval.

$f(-2)=-8-4+12=0$ and $f(3)=-18+6+12=0$: this is Rolle's theorem, $m=0$. $f'(c)=-4c+2=0$ gives $c=\frac{1}{2}$, the midpoint of the roots $-2$ and $3$, the vertex of the parabola.

$f(-2)=-4$, $f(1)=2$, so $m=\frac{6}{3}=2$. $f'(c)=3c^{2}+2c=2$, that is $3c^{2}+2c-2=0$: $c=\frac{-2\pm\sqrt{4+24}}{6}=\frac{-1\pm\sqrt{7}}{3}$, that is $c\approx-1.215$ and $c\approx 0.549$. Both lie in $(-2, 1)$: the theorem promises one, we find two.

$f(0)=0$, $f(2)=16-32-8+30=6$, so $m=3$. $f'(c)=4c^{3}-12c^{2}-4c+15=3$ gives $4c^{3}-12c^{2}-4c+12=0$, which factors by grouping: $4c^{2}(c-3)-4(c-3)=4(c-3)(c^{2}-1)=4(c-3)(c-1)(c+1)$. Three roots, $-1$, $1$ and $3$, but only one in $(0, 2)$: $c=1$. Pitfall: giving all three roots without sorting them.

$f(1)=6$, $f(3)=2$, so $m=\frac{2-6}{2}=-2$. $f'(c)=-\frac{6}{c^{2}}=-2$ gives $c^{2}=3$, and only the positive root lies in $(1, 3)$: $c=\sqrt{3}\approx 1.732$, the geometric mean of the endpoints. Pitfall: the negative slope; $f'(c)=\frac{6}{c^{2}}$ without the minus sign has no solution at all.

$f(-1)=2$, $f(4)=3$, so $m=\frac{1}{5}$. $f'(c)=\frac{1}{2\sqrt{c+5}}=\frac{1}{5}$ gives $\sqrt{c+5}=\frac{5}{2}$, then $c+5=\frac{25}{4}$ and $c=\frac{5}{4}=1.25$.

$f(1)=1$, $f(8)=8^{\frac{2}{3}}=4$, so $m=\frac{3}{7}$. $f'(c)=\frac{2}{3}c^{-\frac{1}{3}}=\frac{3}{7}$ gives $c^{\frac{1}{3}}=\frac{14}{9}$, so $c=\left(\frac{14}{9}\right)^{3}=\frac{2744}{729}\approx 3.764$. Pitfall: $8^{\frac{2}{3}}$ is $\left(\sqrt[3]{8}\right)^{2}=4$, not $\frac{16}{3}$.

$f(1)=0$, $f(4)=\ln 4$, so $m=\frac{\ln 4}{3}$. $f'(c)=\frac{1}{c}=\frac{\ln 4}{3}$ gives $c=\frac{3}{\ln 4}\approx 2.164$.

$f(0)=1$, $f(2)=e^{2}$, so $m=\frac{e^{2}-1}{2}\approx 3.1945$. $f'(c)=e^{c}=m$ gives $c=\ln\left(\frac{e^{2}-1}{2}\right)\approx 1.161$. The logarithm undoes the exponential: $c$ is the logarithm of the slope.

$f(0)=1$, $f(\pi)=-1$, so $m=-\frac{2}{\pi}$. $f'(c)=-\sin c=-\frac{2}{\pi}$ gives $\sin c=\frac{2}{\pi}\approx 0.6366$. On $(0, \pi)$, the sine takes this value twice: $c=\arcsin\frac{2}{\pi}\approx 0.690$ and $c=\pi-\arcsin\frac{2}{\pi}\approx 2.451$. Pitfall: stopping at the calculator value, which only returns the first one; and we work in radians.

If $f$ is CONTINUOUS on the closed interval $[a, b]$, DIFFERENTIABLE on the open interval $\left(a, b\right)$, and $f(a)=f(b)$, then there is at least one $c$ in $\left(a, b\right)$ such that $f'(c)=0$. Note the deliberate asymmetry between the two intervals: continuity is required up to and including the endpoints, differentiability only on the inside. That is what lets you apply the theorem to $\sqrt{1-x^{2}}$ on $[-1, 1]$, whose derivative blows up at both endpoints.

$f$ is a polynomial, so it is continuous and differentiable everywhere: the first two hypotheses hold without discussion. That leaves $f\left(-\sqrt{3}\right)=-3\sqrt{3}+3\sqrt{3}=0$ and $f\left(\sqrt{3}\right)=3\sqrt{3}-3\sqrt{3}=0$: they are equal ✓. Solve $f'(c)=3c^{2}-3=0$, so $c=\pm 1$, and both values lie inside the interval since $\sqrt{3}\approx 1.732$. The figure below marks the two points, $(-1, 2)$ and $(1, -2)$, with their horizontal tangent lines.

You find TWO, while the theorem only promised one. That is exactly what “there exists at least one” means: the theorem gives a lower bound on the number of solutions, never their exact number, and even less their value. An existence statement is deliberately weak, and that is what makes it applicable in a huge variety of situations. Trying to make it say “there exists exactly one” is the most common mistake of the chapter.

$g(1)=1-4+3=0$ and $g(3)=9-12+3=0$ ✓. The theorem gives a $c$ in $\left(1, 3\right)$ with $g'(c)=0$. A parabola is symmetric about its vertex, and its vertex is the only point with a horizontal tangent: $c$ is therefore the midpoint of the two roots, $c=\frac{1+3}{2}=2$. Check: $g'(x)=2x-4$ is indeed zero at $x=2$. For a quadratic function, Rolle's theorem always gives the midpoint, which makes it the simplest case to check that you have understood the statement.

$\sin 0=\sin(2\pi)=0$ ✓, and $\cos c=0$ has two solutions in $\left(0, 2\pi\right)$: $c=\frac{\pi}{2}\approx 1.571$ and $c=\frac{3\pi}{2}\approx 4.712$. You can produce as many solutions as you like by lengthening the interval: on $[0, 2k\pi]$ there are $2k$ of them. The theorem remains true and remains just as terse, which once more illustrates question c.

$f(x)=|x|$ is continuous on $[-1, 1]$ ✓ and $f(-1)=f(1)=1$ ✓. However, it is NOT differentiable at 0, which lies inside the interval: the second hypothesis fails. And the conclusion fails with it, since $f'(x)$ equals $-1$ to the left of 0 and $+1$ to the right, without ever equalling 0. The corner visible in the figure is exactly where the horizontal tangent should have been: the function changes direction without going through a zero slope.

Take $g$ defined on $[0, 1]$ by $g(x)=x$ if $x<1$ and $g(1)=0$. Then $g(0)=g(1)=0$ ✓, $g$ is differentiable on $\left(0, 1\right)$ ✓ with $g'=1$, but it is not continuous at 1. The conclusion fails since $g'$ equals 1 wherever it exists. The function comes back to 0 through a JUMP, and a jump costs no zero slope.

The function $h(x)=x$ on $[0, 1]$ is enough: continuous ✓, differentiable ✓, but $h(0)=0\neq 1=h(1)$. Its derivative equals 1 everywhere and is never zero. It is the cheapest counterexample of the chapter, and it shows that the third hypothesis is the one that carries the whole strength of the theorem: without it, nothing is left.

$m(x)=\frac{1}{x^{2}}$ does satisfy $m(-1)=m(1)=1$, so the hypothesis of equal values at the endpoints holds, which makes the trap believable. But $m$ is not defined at 0, so it is neither continuous nor differentiable on the interval. And $m'(x)=-\frac{2}{x^{3}}$ is never zero: the conclusion is false. It is the most instructive of the three counterexamples, because it satisfies the most visible hypothesis and fails the one students forget to check.

Yes, and it is instructive. The function $p(x)=x^{2}$ on $[-2, 1]$ does not have $p(-2)=p(1)$, since $4\neq 1$: the hypotheses are not all met. Yet $p'(0)=0$ with 0 inside the interval: the conclusion is true. This proves that the implication in Rolle's theorem goes in ONE DIRECTION ONLY. The hypotheses are sufficient, they are not necessary, so you can never argue from a single example that “the conclusion holds, therefore the hypotheses do”.

If $f$ is continuous on $[a, b]$ and differentiable on $\left(a, b\right)$, then there is at least one $c$ in $\left(a, b\right)$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}$. Geometrically: there is at least one point of the curve where the TANGENT line is parallel to the SECANT line joining the two endpoints. When $f(a)=f(b)$, the secant line is horizontal and you get exactly Rolle's theorem.

Slope of the secant line: $\frac{f(4)-f(1)}{4-1}=\frac{2-1}{3}=\frac{1}{3}$. Solve $f'(c)=\frac{1}{2\sqrt{c}}=\frac{1}{3}$, so $2\sqrt{c}=3$, $\sqrt{c}=1.5$ and $c=2.25$. The figure below marks this point, with its dashed tangent line, visibly parallel to the secant line. Notice that $c=2.25$ is not the midpoint of $[1, 4]$, which is 2.5: the promised point has no reason to sit in the middle, except for a parabola.

$\frac{g(3)-g(0)}{3-0}=\frac{(27-9)-0}{3}=6$. Solve $3c^{2}-3=6$, so $c^{2}=3$ and $c=\pm\sqrt{3}$. Only $c=\sqrt{3}\approx 1.732$ belongs to $\left(0, 3\right)$; the value $-\sqrt{3}$ is rejected. This is the check students forget most often: the theorem promises a point IN the open interval, and a solution of the equation that falls outside does not count.

If $f(t)$ is the position of a moving object and $t$ is time, then $\frac{f(b)-f(a)}{b-a}$ is its AVERAGE VELOCITY over the trip, and $f'(c)$ its INSTANTANEOUS velocity at time $c$. The theorem therefore says: on any trip, there is at least one instant when the instantaneous velocity is exactly equal to the average velocity. That is the basis of the average speed camera, which exercise 9 puts to use.

$\frac{h(1)-h(-1)}{1-(-1)}=\frac{1-(-1)}{2}=1$. Solve $3c^{2}=1$, so $c=\pm\frac{1}{\sqrt{3}}\approx\pm 0.577$, and both values lie in $\left(-1, 1\right)$. There is no contradiction with strict increase: it requires $h'\geq 0$, not that $h'$ takes each value only once. Here $h'(x)=3x^{2}$ is a parabola, so it reaches the value 1 twice, once on each side of its minimum. The theorem counts the points where the SLOPE takes a given value, not the points where the function takes it.

The bracket is the equation of the SECANT line joining the points with $x$-coordinates $a$ and $b$. So $g(x)$ is the vertical gap, with its sign, between the curve and its secant line. We have built a new function that measures “how far the curve strays from the straight line”, and it is this gap that we are going to feed to Rolle's theorem.

$g$ is continuous on $[a, b]$ and differentiable on $\left(a, b\right)$ as the difference of $f$, which is by hypothesis, and of a linear function, which always is. At the endpoints: $g(a)=f(a)-f(a)=0$, and $g(b)=f(b)-\left[f(a)+\frac{f(b)-f(a)}{b-a}(b-a)\right]=f(b)-f(b)=0$. So $g(a)=g(b)=0$ ✓. The three hypotheses are met, and it is precisely because we subtracted exactly the secant line that the two values became equal.

Rolle's theorem gives a $c$ in $\left(a, b\right)$ with $g'(c)=0$. Now $g'(x)=f'(x)-\frac{f(b)-f(a)}{b-a}$, since the derivative of the linear part is its slope, a constant. So $g'(c)=0$ reads $f'(c)=\frac{f(b)-f(a)}{b-a}$, which is exactly the conclusion we wanted. The mean value theorem is therefore not one more theorem: it is Rolle's theorem seen in a tilted coordinate system.

$f(3)=9$, $f(1)=1$, the slope of the secant line is $\frac{9-1}{2}=4$, and the secant line is $y=1+4(x-1)=4x-3$. So $g(x)=x^{2}-4x+3$, which is the parabola of exercise 1, question d. It is zero at 1 and at 3 ✓, and its extremum is at $c=2$, where $g(2)=-1$: the largest gap between the curve and its secant line is 1, reached at the midpoint. Final check: $f'(2)=4$, which is indeed the slope of the secant line ✓.

Because Rolle's theorem itself gives no way of computing its point: its proof relies on the Extreme Value Theorem, which states that a function continuous on a closed bounded interval reaches its maximum, without saying where. The whole chain is made of EXISTENCE arguments. In exercises, you find $c$ by solving an equation, but that is extra work, specific to the chosen function, and it fails as soon as the equation $f'(c)=k$ cannot be solved by hand. The theorem remains true in that case, and that is precisely what makes it valuable.

Let $x_{1}<x_{2}$ be two points of the interval. On $[x_{1}, x_{2}]$, $f$ is continuous and differentiable, so the mean value theorem gives a $c$ such that $f(x_{2})-f(x_{1})=f'(c)(x_{2}-x_{1})$. Since $f'(c)=0$ by hypothesis, we get $f(x_{2})=f(x_{1})$. The two points being arbitrary, $f$ takes the same value everywhere: it is constant. Notice that this result, which looks obvious, has NO elementary proof without the theorem.

Apply a to the function $h=f-g$, whose derivative $h'=f'-g'$ is zero on the interval: $h$ is therefore constant, that is, $f=g+k$. This is the foundation of the chapter on ANTIDERIVATIVES: two antiderivatives of the same function on an interval differ by a constant, which is why you always write “plus $C$”. Without this theorem, nothing would guarantee that you have found them all.

It is essential. Take $f(x)=\frac{|x|}{x}$ on $\mathbb{R}\setminus\{0\}$, which equals $-1$ for $x<0$ and $+1$ for $x>0$. Its derivative is zero at every point of its domain, and yet $f$ is not constant. The reason is that $\mathbb{R}\setminus\{0\}$ is not an interval: between a negative point and a positive point, you cannot apply the mean value theorem, since the function is not defined at 0. That is also why the antiderivative of $\frac{1}{x}$ is written $\ln|x|+C$ with a constant that can DIFFER on the two half-lines.

Let $F(x)=\arctan x+\arctan\left(\frac{1}{x}\right)$. Then $F'(x)=\frac{1}{1+x^{2}}+\frac{-\frac{1}{x^{2}}}{1+\frac{1}{x^{2}}}=\frac{1}{1+x^{2}}-\frac{1}{x^{2}+1}=0$ for every $x>0$, after multiplying the numerator and denominator of the second term by $x^{2}$. Since $\left(0, +\infty\right)$ is an interval, $F$ is constant there, and you identify the constant at a convenient point: $F(1)=\frac{\pi}{4}+\frac{\pi}{4}=\frac{\pi}{2}$. So $F(x)=\frac{\pi}{2}$ for every $x>0$. Check at $x=3$: $\arctan 3\approx 1.2490$ and $\arctan\left(\frac{1}{3}\right)\approx 0.3217$, whose sum is $1.5708\approx\frac{\pi}{2}$ ✓.

No: on $\left(-\infty, 0\right)$ the computation of $F'$ is identical and again gives 0, so $F$ is constant there too, but nothing forces the two constants to agree. Evaluate it at $x=-1$: $F(-1)=-\frac{\pi}{4}-\frac{\pi}{4}=-\frac{\pi}{2}$. The two constants are opposites. This is exactly the situation of question c: the domain $\mathbb{R}\setminus\{0\}$ is made of TWO intervals, and the theorem does not carry over from one to the other. Concluding “$F$ is constant on its domain” would be wrong, and that is the mistake the question was trying to provoke.