Mean Value Theorem (MVT) – AP Calculus

The Mean Value Theorem (MVT) is one of the most important theorems in calculus. It guarantees that for a smooth function on an interval, there is at least one point where the instantaneous rate of change equals the average rate of change over the entire interval. In other words, somewhere between $a$ and $b$, the tangent line is parallel to the secant line.

🎯 In this section you will learn

The statement of the Mean Value Theorem
The geometric interpretation: at some point, the tangent line is parallel to the secant line
How to find the $c$ values guaranteed by the MVT
Applications of the MVT (proving inequalities, function behavior)

📌 The Mean Value Theorem (MVT)

$ \text{If } f \text{ is continuous on } [a,b] \text{ and differentiable on } (a,b), \text{ then there exists } c \in (a,b) \text{ such that} $$ $$ f'(c) = \frac{f(b) - f(a)}{b - a} $$ The instantaneous rate of change equals the average rate of change at some point $c

💡 Geometric Interpretation

The secant line connecting $(a, f(a))$ and $(b, f(b))$ has slope $\frac{f(b)-f(a)}{b-a}$. The MVT says there is at least one point $c$ where the tangent line has the same slope — so the tangent line is parallel to the secant line.

📊 Visualizing the MVT

Interactive graph below. The blue curve is the function, the orange line is the secant line between the endpoints, and the green line is the tangent line at the point where slopes match:

Example 1Applying the MVT

Verify that $f(x) = x^2$ satisfies the MVT on $[-1, 3]$. Find all $c$ values that satisfy the theorem.

①

Check conditions

$f(x)=x^2$ is a polynomial → continuous on $[-1,3]$ and differentiable on $(-1,3)$.

②

Compute average slope

$\frac{f(3)-f(-1)}{3-(-1)} = \frac{9 - 1}{4} = \frac{8}{4} = 2$

③

Set $f'(c) = 2$

$f'(x) = 2x$, so $2c = 2$ ⇒ $c = 1$

④

Verify $c$ is in interval

$c = 1$ is in $(-1, 3)$ — valid.

Example 2MVT with a Non-Polynomial

Find all $c$ values for $f(x) = \sqrt{x}$ on $[0, 4]$.

①

Check conditions

$f(x)=\sqrt{x}$ is continuous on $[0,4]$ and differentiable on $(0,4)$.

②

Average slope

$\frac{f(4)-f(0)}{4-0} = \frac{2 - 0}{4} = \frac{1}{2}$

③

Set $f'(c) = 1/2$

$f'(x) = \frac{1}{2\sqrt{x}}$, so $\frac{1}{2\sqrt{c}} = \frac{1}{2}$ ⇒ $\sqrt{c} = 1$ ⇒ $c = 1$

④

Verify

$c = 1$ is in $(0, 4)$ — valid.

Example 3When MVT Does NOT Apply

Does the MVT apply to $f(x) = |x|$ on $[-1, 2]$? Why or why not?

$f(x) = |x|$ is continuous on $[-1, 2]$.
But $f$ is not differentiable at $x = 0$ (corner).
The MVT requires differentiability on the open interval $(-1, 2)$ — since $f$ is not differentiable at $0$, the theorem does NOT apply.

📌 Applications of the MVT

The Mean Value Theorem is used to prove important results in calculus, including:

If $f'(x) = 0$ for all $x$ in an interval, then $f$ is constant on that interval.
If $f'(x) = g'(x)$ for all $x$ in an interval, then $f(x) = g(x) + C$ (they differ by a constant).
If $f'(x) > 0$ on an interval, then $f$ is increasing on that interval.
If $f'(x) < 0$ on an interval, then $f$ is decreasing on that interval.

Example 4Proving a Function is Constant

Show that if $f'(x) = 0$ for all $x$ in $(a,b)$, then $f$ is constant on $(a,b)$.

①

Pick any two points $x_1, x_2$ in $(a,b)$

Assume $x_1 < x_2$. $f$ is differentiable on $(x_1, x_2)$ and continuous on $[x_1, x_2]$.

②

Apply MVT on $[x_1, x_2]$

There exists $c \in (x_1, x_2)$ such that $f'(c) = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$.

③

Since $f'(c) = 0$

$\frac{f(x_2) - f(x_1)}{x_2 - x_1} = 0$ ⇒ $f(x_2) = f(x_1)$

④

Conclusion

$f$ has the same value at any two points → $f$ is constant.

Example 5Proving an Inequality

Use the MVT to prove that $\ln x < x - 1$ for $x > 1$.

①

Let $f(t) = \ln t$ on $[1, x]$ where $x > 1$

$f$ is continuous and differentiable on $(1, x)$.

②

Apply MVT

There exists $c \in (1, x)$ such that $\frac{\ln x - \ln 1}{x - 1} = \frac{1}{c}$

③

Since $c > 1$, we have $\frac{1}{c} < 1$

$\frac{\ln x}{x - 1} < 1$ ⇒ $\ln x < x - 1$

⚠️ Common Mistakes

Forgetting to check conditions: MVT requires continuity on $[a,b]$ AND differentiability on $(a,b)$.
Misunderstanding the conclusion: MVT guarantees at least one $c$, not necessarily exactly one.
Using MVT where it doesn't apply: If the function is not differentiable or not continuous, MVT cannot be used.
Confusing MVT with Rolle's Theorem: Rolle's Theorem is a special case of MVT where $f(a) = f(b)$.

🔍 Key Takeaways

MVT: If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then $f'(c) = \frac{f(b)-f(a)}{b-a}$ for some $c \in (a,b)$.
Geometrically: There is a point where the tangent line is parallel to the secant line.
Rolle's Theorem is a special case: when $f(a) = f(b)$, there is a point with $f'(c) = 0$.
MVT is used to prove properties of functions (constant, increasing, inequalities).
Always verify the conditions before applying the theorem.

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