Applications of the Mean Value Theorem
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 4.4: “The Mean Value Theorem” |
| Direct link | https://openstax.org/books/calculus-volume-1/pages/4-4-the-mean-value-theorem |
| Textbook used in class | Stewart, Calculus, Section 3.2: “The Mean Value Theorem” |
Opening Scenario
A car enters a highway on-ramp at time $t = 0$ and arrives at an exit $100$ miles away at time $t = 1$ hour. The average speed for the trip is $100$ miles per hour. Did the car ever travel at exactly $100$ miles per hour during the trip?
The answer is yes -- and not just by luck. The Mean Value Theorem guarantees it: at some moment during the trip, the car’s instantaneous speed equaled its average speed over the whole trip. That moment might have come during a merge, a stretch of open highway, or a gentle deceleration, but it had to happen.
This is one of the clearest real-world readings of the theorem. The following concepts make it precise.
Quick Reference
Mean Value Theorem (MVT). If $f$ satisfies:
- $f$ is continuous on $[a, b]$,
- $f$ is differentiable on $(a, b)$,
then there exists at least one $c \in (a, b)$ such that $$f'(c) = \frac{f(b) - f(a)}{b - a}.$$
The right side is the average rate of change (slope of the secant line). The left side is the instantaneous rate of change (slope of the tangent line) at $c$. The MVT says those two slopes are equal at some interior point.
How Rolle’s Theorem relates. If additionally $f(a) = f(b)$, the secant slope is zero, and the MVT reduces to Rolle’s Theorem.
Key Concepts
1. Geometric Meaning
Draw the graph of a smooth curve from $(a, f(a))$ to $(b, f(b))$. Draw the secant line connecting those two endpoints. The MVT says there is at least one point $c$ where the tangent line to the curve is exactly parallel to that secant line -- the same slope, though shifted.
Intuitively: if you drive from city $A$ to city $B$ at an average speed of $60$ mph, at some exact moment your speedometer must read $60$ mph. It does not have to read $60$ constantly; it just must hit that value at least once.
2. Applying the MVT: Finding $c$
Example 1. Verify the MVT applies to $f(x) = x^3$ on $[0, 2]$, and find all $c$.
Verification: $f$ is a polynomial, so continuous on $[0, 2]$ and differentiable on $(0, 2)$.
Average rate of change: $$\frac{f(2) - f(0)}{2 - 0} = \frac{8 - 0}{2} = 4.$$
Find $c$: Set $f'(c) = 4$. Since $f'(x) = 3x^2$: $$3c^2 = 4 \Rightarrow c^2 = \frac{4}{3} \Rightarrow c = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}.$$
Check: $\frac{2\sqrt{3}}{3} \approx 1.155 \in (0, 2)$.
Boxed answer: $c = \dfrac{2\sqrt{3}}{3}$.
Recap. The secant slope over $[0, 2]$ is $4$. At $x \approx 1.155$ the tangent to $y = x^3$ also has slope $4$. If you imagine a straight line sliding parallel to the secant, it first touches the curve at exactly that point.
3. Proving Inequalities with the MVT
The MVT is a powerful tool for converting information about derivatives into information about function values. The standard argument:
If $f'(x) \leq M$ on $(a, b)$, then for any $x \in (a, b)$ the MVT gives a $c$ with $f(x) - f(a) = f'(c)(x - a) \leq M(x - a)$. So $f(x) \leq f(a) + M(x - a)$.
Example 2. Show that $\sin x \leq x$ for all $x \geq 0$.
Let $g(x) = x - \sin x$. Then $g(0) = 0$ and $g'(x) = 1 - \cos x \geq 0$ for all $x$ (since $\cos x \leq 1$).
By the MVT applied on $[0, x]$ for any $x > 0$: there exists $c \in (0, x)$ with $$g(x) - g(0) = g'(c) \cdot x.$$ Since $g'(c) \geq 0$ and $x > 0$, the right side is non-negative. So $g(x) \geq g(0) = 0$, which means $x - \sin x \geq 0$, i.e., $\sin x \leq x$.
Boxed answer: $\sin x \leq x$ for all $x \geq 0$. Equality holds only at $x = 0$.
Recap. The key move is defining a function whose sign captures the inequality you want, checking its derivative’s sign, and applying the MVT to connect the two.
4. The Speed Limit Argument
A closely related use: if a derivative is bounded, the function cannot change too fast.
Example 3. Suppose $f(0) = 3$ and $|f'(x)| \leq 5$ for all $x \in [0, 4]$. How large can $f(4)$ be?
By the MVT applied on $[0, 4]$: $$f(4) - f(0) = f'(c) \cdot 4$$ for some $c \in (0, 4)$. Since $|f'(c)| \leq 5$: $$|f(4) - 3| = |f'(c)| \cdot 4 \leq 5 \cdot 4 = 20.$$
So $-20 \leq f(4) - 3 \leq 20$, giving $-17 \leq f(4) \leq 23$.
Boxed answer: $f(4)$ is at most $23$ and at least $-17$.
Recap. This is literally a speed-limit argument: if your speed cannot exceed $5$ for $4$ hours, your displacement cannot exceed $20$.
5. Consequences: Functions with Zero Derivative
The MVT has a corollary that appears constantly in later work.
Corollary. If $f'(x) = 0$ for every $x$ in an interval $(a, b)$, then $f$ is constant on $(a, b)$.
Proof. Take any two points $x_1 < x_2$ in $(a, b)$. The MVT on $[x_1, x_2]$ gives $c$ with $f(x_2) - f(x_1) = f'(c)(x_2 - x_1) = 0 \cdot (x_2 - x_1) = 0$. So $f(x_2) = f(x_1)$. Since $x_1, x_2$ were arbitrary, $f$ is constant.
This result is used in the antiderivative uniqueness theorem: any two antiderivatives of $f$ differ by a constant, because their difference has derivative zero everywhere.
“If $f'(c) = 0$ for one point $c$, then $f$ is constant.” No. The corollary requires $f'(x) = 0$ at every point in the interval, not just one. The derivative being zero at a single point only tells you about the tangent at that point.
6. Uniqueness of Functions with Equal Derivatives
Corollary. If $f'(x) = g'(x)$ for all $x \in (a, b)$, then $f(x) = g(x) + C$ for some constant $C$.
Proof. Let $h(x) = f(x) - g(x)$. Then $h'(x) = f'(x) - g'(x) = 0$. By the corollary above, $h$ is constant, say $h(x) = C$. So $f(x) = g(x) + C$.
This is why the general antiderivative includes the $+ C$: two functions with the same derivative must be the same function shifted vertically. No other difference is possible.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Using the MVT without checking continuity | Applying MVT to $1/x$ on $[-1,1]$ | $1/x$ is not continuous on $[-1,1]$; the theorem does not apply |
| Forgetting to verify $c$ is in the open interval | Finding $c = 0$ for MVT on $[0, 2]$ | $c$ must be in $(0, 2)$; check $c$ is strictly between the endpoints |
| Thinking $c$ is always at the midpoint | Claiming $c = (a+b)/2$ always | $c$ is found by solving $f'(c) = (f(b)-f(a))/(b-a)$; it is rarely the midpoint |
| Applying the zero-derivative corollary at one point | Saying $f$ is constant because $f'(2) = 0$ | The corollary requires $f' = 0$ on the entire interval, not at one point |
Leveled Practice
Level 1 -- Direct Application
Problem 1. Verify the MVT applies to $f(x) = x^2 + x$ on $[1, 3]$, and find all values of $c$.
Show answer
$f$ is a polynomial: continuous on $[1, 3]$, differentiable on $(1, 3)$.
Average rate of change: $\dfrac{f(3) - f(1)}{3 - 1} = \dfrac{12 - 2}{2} = 5$.
Set $f'(c) = 5$: $f'(x) = 2x + 1$, so $2c + 1 = 5 \Rightarrow c = 2$.
Boxed answer: $c = 2 \in (1, 3)$.
Problem 2. Does the MVT apply to $f(x) = |x - 1|$ on $[0, 3]$? If not, which hypothesis fails?
Show answer
$f$ is continuous on $[0, 3]$ (absolute value is continuous). However, $f'(1)$ does not exist (corner at $x = 1$). The differentiability hypothesis fails on the open interval $(0, 3)$.
Boxed answer: The MVT does not apply; differentiability on $(0, 3)$ fails at $x = 1$.
Level 2 -- Multiple Steps
Problem 3. Suppose $f(2) = 5$ and $f'(x) \leq 3$ for all $x \in [2, 6]$. What is the largest possible value of $f(6)$?
Show answer
By the MVT on $[2, 6]$, there exists $c \in (2, 6)$ with $f(6) - f(2) = f'(c) \cdot 4$.
Since $f'(c) \leq 3$: $f(6) - 5 \leq 3 \cdot 4 = 12$, so $f(6) \leq 17$.
This bound is achieved if $f'(x) = 3$ on all of $[2, 6]$, giving $f(6) = 5 + 12 = 17$.
Boxed answer: The largest possible value of $f(6)$ is $17$.
Problem 4. Show that $e^x \geq 1 + x$ for all $x \geq 0$.
Show answer
Let $g(x) = e^x - 1 - x$. Then $g(0) = 1 - 1 - 0 = 0$ and $g'(x) = e^x - 1$.
For $x \geq 0$: $e^x \geq e^0 = 1$, so $g'(x) = e^x - 1 \geq 0$.
By the MVT on $[0, x]$ for any $x > 0$: $g(x) - g(0) = g'(c) \cdot x \geq 0$.
So $g(x) \geq 0$, i.e., $e^x - 1 - x \geq 0$, i.e., $e^x \geq 1 + x$.
Boxed answer: $e^x \geq 1 + x$ for all $x \geq 0$.
Level 3 -- Deeper Problems
Problem 5. Two runners start a race at the same time and finish at the same time. Must there be a moment during the race when both runners have the same instantaneous speed? (Assume both runners’ positions are differentiable functions of time.)
Show answer
Not necessarily. Consider the function $h(t) = p_1(t) - p_2(t)$, the difference in positions. At the start and end, $h = 0$, so by Rolle’s Theorem there is a $c$ where $h'(c) = 0$, meaning $p_1'(c) = p_2'(c)$: their speeds are equal at time $c$.
Wait -- this is exactly the conclusion: yes, there must be such a moment. The argument is to apply Rolle’s Theorem to the position difference, which starts and ends at zero.
Boxed answer: Yes. The difference in positions is zero at start and finish; Rolle’s Theorem guarantees a moment when the rates of change are equal, i.e., the runners have the same speed.
Mastery Checklist
Mental Model
The MVT is the average-matches-instantaneous guarantee. Whatever your average rate of change was over an interval, the instantaneous rate must have equaled that average at some moment inside the interval. The secant line across the interval has a certain slope; somewhere in the middle, the tangent line has the exact same slope.
The applications all follow one pattern: turn information about the derivative (bounded, zero, equal to another derivative) into a conclusion about the function values (bounded change, constant, equal up to a constant). The MVT is the bridge from rate of change to function change.
Connections
Within Chapter 3
- Rolle’s Theorem (Section 3.2): The MVT is the generalization of Rolle. Rolle handles the special case where the secant slope is zero; the MVT handles the general slope.
- Increasing/Decreasing Test (Section 3.3): The proof that $f' > 0 \Rightarrow f$ is increasing uses exactly the MVT: on any subinterval $[x_1, x_2]$, $f(x_2) - f(x_1) = f'(c)(x_2-x_1) > 0$.
- Antiderivatives (Section 3.9): The uniqueness of antiderivatives up to a constant follows directly from the zero-derivative corollary.
Toward MATH162
- Integration (Chapter 4): The connection between antiderivatives and definite integrals (the Fundamental Theorem of Calculus) relies on the MVT in its proof.
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