Rolle’s Theorem and the Mean Value Theorem

An average rate must occur somewhere

If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ , the mean value theorem guarantees an interior $c$ with

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

Geometrically, one tangent is parallel to the endpoint secant. Physically, a differentiable position function has an instantaneous velocity equal to its average velocity somewhere in the interval. The theorem does not say when this happens or that it happens only once.

Continuity is required at both endpoints as well as inside. Differentiability is required only in the open interval; endpoint derivatives are unnecessary.

Visual guide

VISUAL GUIDEA tangent parallel to the secant

x²
Secant y = 2x
Tangent y = 2x − 1

For f(x) = x² on [0, 2], the secant has slope 2. At c = 1 the derivative 2c also equals 2. The orange tangent and dashed secant are parallel, illustrating the Mean Value Theorem.

Rolle’s special case

When $f(a)=f(b)$ , the average slope is zero and Rolle's theorem gives $f'(c)=0$ . To see why, a continuous function on a closed interval attains extrema. If it is not constant and its endpoint heights agree, some nontrivial extremum occurs inside. Differentiability then forces zero slope there. The general MVT follows by subtracting the secant line and applying Rolle's theorem to the difference.

Worked example: locating a guaranteed rate

For $f(x)=x^3$ on $[0,2]$ , the average slope is $(8-0)/2=4$ . Since $f'(x)=3x^2$ , solve $3c^2=4$ . The only solution in $(0,2)$ is $c=2/\sqrt3$ . The negative root is irrelevant because it lies outside the theorem's interval.

Worked example: why a corner invalidates the conclusion

For $f(x)=|x|$ on $[-1,1]$ , endpoint values agree. Yet the derivative is $-1$ on the left and $1$ on the right, with no zero derivative inside. Rolle's theorem does not fail: its differentiability hypothesis fails at zero. Listing the hypothesis explicitly prevents a false contradiction.

Consequences that justify graph reasoning

If $f'>0$ throughout an interval, applying the MVT to any two points shows the later value is greater, so $f$ is strictly increasing. If $f'=0$ throughout an interval, the same reasoning shows $f$ is constant. This is an interval statement: a zero derivative at just one point does not imply a constant function.

If $|f'|\le M$ on the interval, then $|f(b)-f(a)|\le M|b-a|$ . This bounds how much an output can change when an input changes. The bound needs to hold throughout the interval, not merely at its endpoints.

Practice

Find the MVT point for $x^2$ on $[1,3]$ .
Can Rolle's theorem be applied to $1/x$ on $[-1,1]$ ?
If a differentiable temperature record has $|T'|\le2$ °C/min, bound its change over three minutes.

Show worked solutions

The secant slope is $(9-1)/2=4$ , so $2c=4$ and $c=2$ .
No. The function is undefined at zero and is not continuous on the closed interval.
The magnitude of the change is at most $2\cdot3=6$ °C, assuming the bound holds across those three minutes.