Mean Value Theorems

The mean value theorems link the average rate of change of a function over an interval to its instantaneous rate at some interior point. They are the bridge between "the derivative is positive" and "the function increases", and they give quick error bounds. The working tasks are finding the point $c$ promised by Rolle's or Lagrange's theorem, checking whether a theorem applies to a given function, and drawing a consequence such as the number of real roots or a numerical bound.

Let $f$ be

Then there is at least one $c \in (a, b)$ with $f'(c) = 0$: somewhere the tangent is horizontal.

Why. By the extreme value theorem $f$ attains a maximum and a minimum on $[a, b]$. If both occur at the endpoints, $f$ is constant and every $c$ works. Otherwise one occurs at an interior point, where Fermat's theorem gives $f'(c) = 0$.

If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, there is $c \in (a, b)$ with

Geometrically, some tangent is parallel to the chord joining $(a, f(a))$ and $(b, f(b))$. Rolle's theorem is the special case $f(a) = f(b)$. Proof: apply Rolle to $f$ minus the chord.

Equivalent forms. $f(b) = f(a) + (b - a)f'(c)$, and with $b = a + h$: $f(a + h) = f(a) + hf'(a + \theta h)$ for some $0 < \theta < 1$.

If $f$ and $g$ are continuous on $[a, b]$, differentiable on $(a, b)$, and $g'(x) \ne 0$ on $(a, b)$, there is $c \in (a, b)$ with

With $g(x) = x$ this is Lagrange's theorem. Cauchy's version is the step used to prove l'Hôpital's rule.

Closed versus open. Continuity is needed on the closed interval $[a, b]$, differentiability only on the open interval $(a, b)$. $\sqrt{x}$ on $[0, 4]$ qualifies even though $f'(0)$ does not exist.

If a hypothesis fails, the conclusion may fail:

The mean value theorem is the first case of Taylor's theorem: if $f$ has $n + 1$ derivatives on an interval around $a$, then

for some $\xi$ between $a$ and $x$ (Lagrange form of the remainder). With $n = 0$ this is exactly Lagrange's theorem.

$f(x) = x^2 - 4x + 3$ on $[1, 3]$: a polynomial, so continuous and differentiable; $f(1) = 0 = f(3)$. $f'(c) = 2c - 4 = 0$ gives $c = 2 \in (1, 3)$.

$f(x) = \sqrt{x}$ on $[0, 4]$: chord slope $(2 - 0)/4 = 1/2$; $f'(c) = 1/(2\sqrt{c}) = 1/2$ gives $c = 1$.

For $f(x) = x^2$ on any $[a, a + h]$: $(a + h)^2 - a^2 = h \cdot 2(a + \theta h)$ gives $\theta = 1/2$, so $c$ is always the midpoint.

$f(x) = x^3 - x$ on $[0, 2]$: chord slope $(6 - 0)/2 = 3$; $3c^2 - 1 = 3$ gives $c = 2/\sqrt{3} \approx 1.155$ (the negative root lies outside the interval).

$f(x) = x^2$, $g(x) = x^3$ on $[1, 2]$: $(4 - 1)/(8 - 1) = 3/7$ must equal $2c/(3c^2) = 2/(3c)$, so $c = 14/9 \approx 1.556 \in (1, 2)$.

Estimate $\sqrt{101}$. By Lagrange, $\sqrt{101} - 10 = 1/(2\sqrt{c})$ with $100 < c < 101$, so

Since $\sqrt{101} < 10.05$, the left side exceeds $1/20.1 \approx 0.04975$, giving $10.0497 < \sqrt{101} < 10.05$ (true value $10.04988$).

$f(x) = x^3 + x - 1$: $f'(x) = 3x^2 + 1 > 0$, so $f$ is strictly increasing and has at most one real root. $f(0) = -1 < 0 < 1 = f(1)$, so by the intermediate value theorem it has exactly one, in $(0, 1)$.

For $x > 0$, $h(x) = \arctan x + \arctan(1/x)$ has $h'(x) = \frac{1}{1 + x^2} - \frac{1}{x^2} \cdot \frac{1}{1 + 1/x^2} = 0$. So $h$ is constant, and $h(1) = \pi/4 + \pi/4 = \pi/2$.

Approximate $e^{0.1}$ by $1 + 0.1 + 0.01/2 = 1.105$. The remainder is $e^{\xi}(0.1)^3/3!$ with $0 < \xi < 0.1$, so the error is at most $e^{0.1} \cdot 0.001/6 < 0.000185$. The actual error is about $0.000171$.

For any real $a < b$, Lagrange gives $\sin b - \sin a = (b - a)\cos c$, and $|\cos c| \le 1$, so $|\sin b - \sin a| \le |b - a|$. The same argument with $f(x) = \ln x$ on $[1, 1 + x]$ gives $\frac{x}{1 + x} < \ln(1 + x) < x$ for $x > 0$, because $f'(c) = 1/c$ lies between $1/(1 + x)$ and 1.