The Mean Value Theorem and Local Approximation

If a journey covers 100 kilometres in two hours, must the instantaneous speed ever equal 50 kilometres per hour? Under a differentiable position model for one-directional travel, the mean value theorem says yes. Its force comes from linking information about a whole interval to a derivative at some interior point.

An interior extremum has a stationary derivative

Suppose $f$ has a local maximum at an interior point $c$ and is differentiable there. For small positive $h$, $[f(c+h)-f(c)]/h\leq0$; for small negative $h$, the quotient is nonnegative. If the two-sided limit exists, it must be both nonpositive and nonnegative, hence zero. A local minimum is similar. This is Fermat’s stationary-point theorem. It does not apply to endpoints or points where the derivative fails to exist, and its converse is false: $x^3$ is stationary at zero without an extremum.

Rolle and the mean value theorem

Rolle’s theorem: if $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, some $c\in(a,b)$ has $f'(c)=0$. EVT gives attained extremes. If the function is nonconstant, at least one extreme differs from the equal endpoint value and is attained inside; Fermat’s theorem applies. If it is constant, every interior derivative is zero.

For the mean value theorem, remove the secant line. Define

$$g(x)=f(x)-f(a)-\frac{f(b)-f(a)}{b-a}(x-a).$$

Its endpoint values are both zero, so Rolle gives an interior $c$ with

$$f'(c)=\frac{f(b)-f(a)}{b-a}.$$

The continuity and differentiability assumptions remain essential. For $f(x)=|x|$ on $[-1,1]$, the endpoint secant slope is zero but no derivative equals zero: slopes are $-1$ and $1$, and the derivative fails at zero.

For $f(x)=x^2$ on $[1,3]$, the secant slope is $4$, matched by $f'(2)=4$. MVT guarantees at least one matching point, not necessarily a unique one.

Consequences that calculus repeatedly uses

If $f'>0$ everywhere on an interval, any two inputs $u<v$ satisfy $f(v)-f(u)=f'(c)(v-u)>0$. Thus $f$ is strictly increasing. If $f'=0$ throughout an interval, the same equation shows it is constant. If $f'=g'$ on an interval, apply the zero-derivative result to $f-g$: the functions differ by a constant.

“On an interval” cannot be dropped. The function $1$ on negative inputs and $2$ on positive inputs has derivative zero throughout its disconnected domain, but is not one constant across that domain. Each connected interval can carry its own constant.

If $|f'|\leq M$ throughout the interval, MVT yields $|f(v)-f(u)|\leq M|v-u|$. This turns a derivative bound into a finite-change bound. For example, $|(\sin x)'|=|\cos x|\leq1$ implies $|\sin v-\sin u|\leq|v-u|$ for radian inputs.

Linearization with an error estimate

The local linear approximation at $a$ is $L(x)=f(a)+f'(a)(x-a)$. For $f(x)=\sqrt{x}$ near $4$, $L(x)=2+(x-4)/4$, so $\sqrt{4.1}\approx2.025$.

If $f$ is twice continuously differentiable on an interval joining $a$ and $x$ and $|f''|\leq M$ there, Taylor’s first-order remainder bound gives

$$|f(x)-L(x)|\leq\frac M2|x-a|^2.$$

One proof follows from the integral remainder after integration is established; a Cauchy mean value theorem argument also proves it. We state the bound here with its extra smoothness assumption rather than treating it as part of the derivative definition. For the square-root example on $[4,4.1]$, $|f''(u)|=1/(4u^{3/2})\leq1/32$, giving error at most $0.1^2/64=0.00015625$. The negative second derivative also shows the tangent lies above this concave-down curve.

Exercises

1. Find a point supplied by MVT for $x^3$ on $[0,2]$.
2. Explain precisely why Rolle’s theorem fails for $|x|$ on $[-1,1]$.
3. Prove that two antiderivatives of $2x$ on $\mathbb R$ differ by a constant.
4. Linearize $\log x$ at $1$ and approximate $\log(1.02)$.
5. Bound that approximation error using $|f''|\leq1$ on $[1,1.02]$.
6. If a differentiable function has $|f'|\leq3$ on $[2,5]$, bound $|f(5)-f(2)|$.

Worked solutions

1. The secant slope is $4$. Solve $3c^2=4$ in $(0,2)$: $c=2/\sqrt3$.
2. It is continuous with equal endpoint values, but not differentiable at the interior point zero.
3. Their difference has derivative zero throughout the interval $\mathbb R$; MVT forces that difference to be constant.
4. $L(x)=x-1$, giving $0.02$.
5. The bound is $(1/2)(0.02)^2=0.0002$.
6. At most $3(5-2)=9$.