The Mean Value Theorem and its uses

What you will be able to do

• state the Mean Value Theorem and explain it with a chord and a tangent;
• find the point $c$ for a given function;
• use the theorem for speed, for inequalities and for constants.

The theorem

A car covers $150$ km in $2$ hours. Its average speed is $75$ km/h. Could the car have been below $75$ km/h all the time? No. And above $75$ km/h all the time? No. At some instant the speedometer read exactly $75$ km/h. This is the Mean Value Theorem, also called Lagrange's mean value theorem.

Let $f$ be continuous on $[a, b]$ and differentiable on $(a, b)$. Then there is a number $c$ in $(a, b)$ with $f'(c) = \dfrac{f(b) - f(a)}{b - a}$.

On the graph, $\dfrac{f(b) - f(a)}{b - a}$ is the slope of the chord joining the two end points. The theorem says that some tangent has the same slope, so it is parallel to the chord.

Rolle's theorem is the case $f(a) = f(b)$, where the chord is horizontal. In fact the Mean Value Theorem follows from Rolle's theorem. Subtract the chord from $f$: the function $g(x) = f(x) - f(a) - \dfrac{f(b) - f(a)}{b - a}(x - a)$ has $g(a) = g(b) = 0$. Rolle gives $g'(c) = 0$, which is the result.

Finding the point $c$

1. Check continuity on $[a, b]$ and differentiability on $(a, b)$.
2. Work out the chord slope $\dfrac{f(b) - f(a)}{b - a}$.
3. Solve $f'(c) =$ the chord slope.
4. Keep the solutions that lie in the open interval $(a, b)$.

For $f(x) = x^2$ on $[1, 3]$, the chord slope is $\dfrac{9 - 1}{2} = 4$. Then $2c = 4$ gives $c = 2$.

What the theorem is used for

• Speed. If a car's average speed is $v$ over a trip, then at some instant its speed was exactly $v$.
• A zero derivative. If $f'(x) = 0$ for every $x$ in an interval, then $f$ is constant there. For any two points $a < b$ the theorem gives $f(b) - f(a) = f'(c)(b - a) = 0$.
• Increasing and decreasing. If $f'(x) > 0$ on an interval, then $f$ is increasing there.
• Inequalities. If you know a bound on $f'$, you get a bound on $f(b) - f(a)$.

An inequality. Show that $\dfrac{x}{1 + x} < \ln(1 + x) < x$ for $x > 0$. Apply the theorem to $f(t) = \ln(1 + t)$ on $[0, x]$. There is a $c$ in $(0, x)$ with

Since $0 < c < x$, the number $\dfrac{1}{1 + c}$ lies between $\dfrac{1}{1 + x}$ and $1$. So $\ln(1 + x)$ lies between $\dfrac{x}{1 + x}$ and $x$.

A note on syllabuses. CBSE and JEE ask the theorem directly. AP Calculus requires it, and expects you to state that $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$. A-Level and IB do not name it.

A worked example

Verify the Mean Value Theorem for $f(x) = x^3 - 2x$ on $[0, 3]$ and find $c$.

1. Conditions. $f$ is a polynomial, so it is continuous and differentiable everywhere.
2. The chord slope. $f(0) = 0$ and $f(3) = 27 - 6 = 21$. The slope is $\dfrac{21 - 0}{3} = 7$.
3. Solve $f'(c) = 7$. $3c^2 - 2 = 7$ gives $c^2 = 3$, so $c = \pm\sqrt{3}$.
4. Keep the right one. Only $c = \sqrt{3} \approx 1.732$ lies in $(0, 3)$.

Practise with new numbers

Try it in Python

The program finds the point $c$ for three functions by halving the interval. It also tests the inequality $\dfrac{x}{1 + x} < \ln(1 + x) < x$ at three values of $x$.

Common mistakes

• Using the midpoint. The point $c$ is the middle of the interval only for a quadratic.
• Accepting a point outside $(a, b)$. Check that each solution lies in the open interval.
• Forgetting the conditions. For $f(x) = \dfrac{1}{x}$ on $[-1, 1]$ the theorem does not apply, because $f$ is not continuous at $0$.
• Reading the theorem as a method. It says that $c$ exists. It does not say where $c$ is.

Remember these

• $f'(c) = \dfrac{f(b) - f(a)}{b - a}$ for some $c$ in $(a, b)$: a tangent parallel to the chord.
• Rolle's theorem is the special case $f(a) = f(b)$.
• Zero derivative on an interval means constant. Positive derivative means increasing.
• Apply the theorem to a well-chosen function to get an inequality.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find $c$ for $f(x) = x^2$ on $[1, 5]$.

2. A van covers $180$ km in $3$ hours. What does the Mean Value Theorem say?

3. Why can the theorem not be applied to $f(x) = \dfrac{1}{x}$ on $[-1, 1]$?

4. If $f'(x) = 0$ for every $x$ in $(a, b)$, what can you say about $f$?

5. Show that $e^x > 1 + x$ for $x > 0$.

References

• OpenStax, Calculus Volume 1, Chapter 4, section on the Mean Value Theorem.
• NCERT, Mathematics Part I for Class XII, Chapter 5, section on the Mean Value Theorem.