Unit 5: Analytical Applications of Differentiation

Mean Value Theorem and Rolle’s Theorem

Derivatives start as a local idea: the derivative at a point gives the slope of the tangent line right there. In this unit, you learn to scale that local information into global conclusions about an entire interval, such as where a function increases or decreases and where it must have peaks, valleys, or horizontal tangents.

The main bridge from local to global is the Mean Value Theorem (MVT). Intuitively, MVT says that if a function is continuous and smooth enough on an interval, then somewhere in that interval the instantaneous rate of change matches the average rate of change over the whole interval. Equivalently: there must be some point where the slope of the tangent line equals the slope of the secant line connecting the endpoints.

Average rate of change vs. instantaneous rate of change

The average rate of change on the interval from x=a to x=b is the slope of the secant line:

\frac{f(b)-f(a)}{b-a}

The instantaneous rate of change at x=c is the derivative:

f'(c)

MVT guarantees these match at least once inside the interval (when its conditions are met).

Mean Value Theorem (MVT)

If f is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists at least one number c in (a,b) such that:

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

Why the conditions matter

Continuity on [a,b] rules out jumps, holes, and vertical asymptotes on the interval. Differentiability on (a,b) rules out corners, cusps, and vertical tangents inside the interval. If either condition fails, the tangent slope that MVT promises may not exist.

A common mistake is applying MVT on an interval that includes a point where the function is not continuous or not differentiable.

Rolle’s Theorem (special case of MVT)

Rolle’s Theorem is the special case where the endpoints have the same function value:

f(a)=f(b)

Then the average slope is:

\frac{f(b)-f(a)}{b-a}=0

So there exists at least one c in (a,b) such that:

f'(c)=0

Interpretation: a continuous, differentiable curve that starts and ends at the same height must have at least one horizontal tangent between those two points.

What MVT lets you prove (beyond finding c)

MVT is often used on AP problems to justify existence or global behavior.

  1. If the derivative is zero everywhere on an interval, then the function is constant there. (If two outputs differed, the average slope would be nonzero, and MVT would force a nonzero derivative somewhere—contradiction.)

  2. If the derivative is positive everywhere on an interval, then the function is increasing on that interval.

f'(x)>0

  1. One-to-one (uniqueness) arguments: if the derivative is positive everywhere on an interval, then the function is strictly increasing there and cannot repeat an output value (so it is one-to-one on that interval).

Example 1: Verifying MVT and finding a valid c

Let f(x)=x^2 on [1,3].

  1. Conditions: polynomials are continuous and differentiable everywhere, so MVT applies.
  2. Average rate of change:

\frac{f(3)-f(1)}{3-1}=\frac{9-1}{2}=4

  1. Derivative:

f'(x)=2x

  1. Set tangent slope equal to secant slope:

2c=4

c=2

Example 2: When MVT does not apply

Let f(x)=|x| on [-1,1].

The function is continuous on [-1,1] but not differentiable at x=0 (a corner).

Average slope:

\frac{f(1)-f(-1)}{1-(-1)}=\frac{1-1}{2}=0

If MVT applied, there would be a c in (-1,1) with:

f'(c)=0

But for x>0,

f'(x)=1

and for x