Using the Mean Value Theorem

3. Visual Connection

5. Concept Development

1. Mean Value Theorem statement

If \(f\) is continuous on the closed interval \([a,b]\) and differentiable on the open interval \((a,b)\), then at least one number \(c\in(a,b)\) satisfies

The right side is the average rate of change over the whole interval. The left side is an instantaneous rate at an interior input.

2. Geometric meaning

The endpoint quotient is the slope of the secant line through \((a,f(a))\) and \((b,f(b))\). The theorem guarantees at least one interior tangent line parallel to that secant line.

3. Why both hypotheses matter

Continuity on \([a,b]\) prevents breaks, holes, and jumps across the interval. Differentiability on \((a,b)\) prevents interior corners, cusps, vertical tangents, and other points where a finite derivative is unavailable. Endpoints need continuity but do not need two-sided derivatives.

4. A complete solution workflow

1. State why \(f\) is continuous on \([a,b]\).
2. State why \(f\) is differentiable on \((a,b)\).
3. Compute the average slope \(m=(f(b)-f(a))/(b-a)\).
4. Solve \(f'(c)=m\).
5. Keep only solutions in the open interval \((a,b)\).

5. Existence does not mean uniqueness

The theorem guarantees at least one value of \(c\). A function may have one, several, or infinitely many points whose tangent slope equals the secant slope. Solving the derivative equation finds all candidates; the theorem alone does not locate them.

6. Rolle's Theorem

When \(f(a)=f(b)\), the secant slope is zero. The Mean Value Theorem becomes Rolle's Theorem:

Its geometric conclusion is an interior horizontal tangent.

7. A theorem can fail to apply without proving the conclusion false

If a hypothesis fails, the correct statement is that the theorem provides no guarantee. An appropriate \(c\) might still exist by coincidence. To disprove a claimed application, identify the failed hypothesis; to disprove the conclusion, also show that no suitable \(c\) exists.

8. Motion interpretation

For a differentiable position function \(s(t)\),

is average velocity, while \(s'(c)\) is instantaneous velocity. The theorem says that during the trip, the instantaneous velocity equals the trip's average velocity at least once. It does not claim that the equality occurs at the midpoint in time.

9. Using endpoint data from tables

An explicit formula is unnecessary for the existence conclusion. If continuity and differentiability are given and a table provides \(f(a)\) and \(f(b)\), those endpoint values determine the guaranteed derivative value. A formula for \(f'\) is needed only when the problem asks for the actual input \(c\).

10. Bounding total change with derivative bounds

If \(m\le f'(x)\le M\) throughout \((a,b)\), the Mean Value Theorem gives

This converts bounds on every instantaneous rate into bounds on the total change.

11. Important consequences

If \(f'(x)=0\) throughout an interval, then \(f\) is constant there. More generally, if \(f'(x)=g'(x)\) throughout an interval, then \(f(x)-g(x)\) is constant. These conclusions follow by applying the theorem between any two inputs in the interval.

12. Proof idea through Rolle's Theorem

Subtract the endpoint secant line from \(f\):

Then \(h(a)=h(b)=0\). Rolle's Theorem guarantees \(h'(c)=0\), which rearranges to the Mean Value Theorem equation.

13. Common errors

• Checking differentiability but forgetting continuity on the closed interval.
• Using endpoint derivatives instead of endpoint function values in the average slope.
• Claiming that \(c=(a+b)/2\) without solving the derivative equation.
• Keeping a solution equal to an endpoint even though \(c\) must lie in \((a,b)\).
• Claiming a unique \(c\) when the theorem guarantees only existence.
• Confusing this theorem with the Mean Value Theorem for Integrals.

6. Detailed Worked Example and Error Check

Example 1: A polynomial. Apply the theorem to \(f(x)=x^2\) on \([1,4]\). A polynomial is continuous and differentiable everywhere, so both hypotheses hold.

Solving \(2c=5\) gives \(\boxed{c=5/2}\), which lies in \((1,4)\).

Example 2: Rolle's Theorem with multiple points. For \(f(x)=\sin x\) on \([0,2\pi]\), continuity and differentiability hold and \(f(0)=f(2\pi)=0\).

The interior solutions are \(\boxed{c=\pi/2}\) and \(\boxed{c=3\pi/2}\). The theorem guarantees at least one; solving identifies both.

Example 3: A rational function. Apply the theorem to \(f(x)=1/x\) on \([1,4]\). The interval contains no zero, so \(f\) is continuous on \([1,4]\) and differentiable on \((1,4)\).

Solving \(-1/c^2=-1/4\) gives \(c=\pm2\), but only \(\boxed{c=2}\) belongs to \((1,4)\).

Example 4: A failed hypothesis. Let \(f(x)=|x|\) on \([-1,1]\). The function is continuous and has equal endpoint values, but it is not differentiable at \(x=0\). Moreover, \(f'(x)=-1\) for \(x<0\) and \(f'(x)=1\) for \(x>0\), so no interior point has derivative zero. This shows why differentiability is essential.

Example 5: Motion. A vehicle's differentiable position satisfies \(s(2)=12\) meters and \(s(7)=52\) meters. Its average velocity is

If \(s\) is continuous on \([2,7]\) and differentiable on \((2,7)\), some \(c\in(2,7)\) satisfies \(\boxed{s'(c)=8\text{ m/s}}\).

Example 6: Endpoint data only. Suppose \(f\) is continuous on \([1,5]\), differentiable on \((1,5)\), \(f(1)=3\), and \(f(5)=19\). Then

Therefore some \(c\in(1,5)\) satisfies \(\boxed{f'(c)=4}\), even though the exact value of \(c\) cannot be found from the endpoint data alone.

Example 7: Bounding change. Suppose \(2\le f'(x)\le5\) for \(1

Thus \(6\le f(4)-7\le15\), so \(\boxed{13\le f(4)\le22}\).