The **mean value theorem** (MVT), also known as **Lagrange's mean value theorem (LMVT)**, provides a formal framework for a fairly intuitive statement relating change in a function to the behavior of its derivative. The theorem states that the derivative of a continuous and differentiable function must attain the function's average rate of change (in a given interval). For instance, if a car travels 100 miles in 2 hours, then it must have had the exact speed of 50 mph at some point in time.

Mean Value TheoremSuppose that a function \(f\) is

- continuous on the closed interval \([a,b],\) and
- differentiable on the open interval \((a,b).\)
Then, there is a number \(c\) such that \(a<c<b\) and \(f'(c)=\frac{f(b)-f(a)}{b-a}.\)

Simple-sounding as it is, the mean value theorem actually lies at the heart of the proof of the fundamental theorem of calculus, and is itself based ultimately on properties of the real numbers. There is a slight generalization known as Cauchy's mean value theorem; for a generalization to higher derivatives, see Taylor's theorem.

#### Contents

- Interpretation
- Proof
- Examples
- Mean Value Theorem for Integrals
- Other Applications
- See Also

## Interpretation

The statement seems reasonable upon inspection of an example or two. Below, \(f'(c)\) is the slope of the tangent lines in the interval \((a,b)\), and \( \frac{f(b)-f(a)}{b-a}\) is the slope of the secant line joining the two endpoints \(\big(a, \, f(a)\big)\) and \(\big(b, \, f(b)\big)\).

Note that the mean value theorem does not restrict \(c\) to only one value, nor does it tell us where \(c\) is (other than inside the interval).

Note also that \(f\) is not required to be differentiable at the end points. Two examples suffice to illustrate it: \(f(x)=\sqrt{1-x^2}\) on \([-1, 1]\) and \(f(x)=\arcsin x\) on \([-1, 1]\).

## Proof

The mean value theorem is best understood by first studying the restricted case known as Rolle's theorem.

Rolle's TheoremSuppose that a function \(f\) is continuous on \([a, b]\), differentiable on \((a, \, b)\), and that \(f(a) = f(b)\). Then, there is a number \(c\) such that \(a<c<b\) and \(f'(c) = 0\).

In other words, if a function has the same value at two points, then it must "level" somewhere between those points. By considering whether the function is increasing or decreasing immediately after the first point, it becomes clear that neither option can continue indefinitely if the function is to return to the same value; therefore, there must be a local maximum or minimum before the next point occurs.

Rolle's theorem quickly turns into the mean value theorem by simply skewing the graph of the function.

Let all be as in the theorem statement above.

Define a new function \(h\) as the difference between \(f\) and the line passing through points \(\big(a, f(a)\big)\) and \(\big(b, f(b)\big)\). This line has equation

\[y-f(a)=\dfrac{f(b)-f(a)}{b-a} (x-a) \implies y= f(a)+\dfrac{f(b)-f(a)}{b-a} (x-a).\]

Therefore, the function \(h\) has equation

\[h(x)=f(x)- f(a)-\dfrac{f(b)-f(a)}{b-a} (x-a).\]

Next, Rolle's theorem is useful. The function \(h\) satisfies the conditions of the theorem:

- The function \(h\) is continuous on \([a,b]\) because it is the sum of \(f\) and a first-degree polynomial, both of which are continuous.
The function is differentiable on \((a,b)\) because both \(f\) and the first-degree polynomial are differentiable. In fact, we can compute it directly: \[ h'(x)= f'(x)-\frac{f(b)-f(a)}{b-a} .\]

\(h(a)=h(b).\)

By Rolle's theorem, there exists a value \(c\) in \((a,b)\) such that \(h'(c)=0\). Therefore,

\[h'(c)=f'(c)-\dfrac{f(b)-f(a)}{b-a} =0 \implies f'(c)=\dfrac{f(b)-f(a)}{b-a}.\ _\square\]

\(f(x) \) is a differentiable function that satisfies \(5 \leq f'(x) \leq 14\) for all \(x\). Let \(a\) and \(b\) be the maximum and minimum values, respectively, that \(f(11) - f(3) \) can possibly have, then what is the value of \(a+b\)?

## Examples

Determine all the numbers \(c\) that satisfy the conclusion of the mean value theorem for \(f(x)=x^{3}+2x^{2}+x\) on the interval \([-2,2]\).

Observe that \(f(x)\) is a polynomial which is continuous and differentiable over any interval, so the mean value theorem applies. Indeed, we can compute that \(f'(x)=3x^2+4x+1\), and the average rate of change is

\[ \frac{f(b)-f(a)}{b-a} = \frac{f(2)-f(-2)}{2-(-2)} = \frac{18 - (-2)}{4} = 5.\]

We want to find the values of \(c\) such that \(f'(c)=5:\)

\[\begin{align}3c^{2}+4c+1&=5\\3c^{2}+4c-4&=0\\(c+2)(3c-2)&=0\\c&=-2, ~ \frac{2}{3}.\end{align}\]

Note that \(c=-2\) is one of the endpoints, so the mean value theorem actually guarantees another \(c\) in the interior, and indeed we have \(c=\frac{2}{3}\). \(_\square\)

A car starts from rest and drives a distance of \(10 \text{ km}\) in \(30 \text{ min}\). Use the mean value theorem to show that the car attains a speed of \(20 \text{ km/hr}\) at some point(s) during the interval.

The mean value theorem says that the average speed of the car (the slope of the secant line) is equal to the instantaneous speed (slope of the tangent line) at some point(s) in the interval.

The average velocity is

\[\frac{\Delta y}{\Delta x}=\frac{10 \text{ km}-0}{0.5 \text{ hr}-0}=20 \text{ km/hr}.\]

From the mean value theorem, we can find \(c\) in the interval such that

\[f'(c)=\frac{dx}{dt}=20 \text{ km/hr}. \ _\square\]

Suppose that \(f(x)\) is a differentiable function for all \(x\). If \(f'(x)\leq 7\) for all \(x\) and \(f(2)=-4\), what is the maximum value of \(f(5)?\)

Since \(f(x)\) is differentiable on all intervals, we can choose any two points. So from the mean value theorem, we have

\[\begin{align}\frac{f(5)-f(2)}{5-2}=f'(c)&\leq 7\\\frac{f(5)-(-4)}{3}&\leq 7\\f(5)&\leq 17.\end{align}\]

So the maximum possible value of \(f(5)\) is \(17\). \(_\square\)

Given that \(f(x)\) is an arbitrary quadratic polynomial:

\[f(x)=Kx^{2}+Lx+M \quad (K\neq 0),\]

show that the point \(c\) whose existence is guaranteed by the mean value theorem is the mid-point of the interval \([a,b]\).

From the mean value theorem, we have

\[\begin{align}f'(c)&=\frac { f(b)-f(a) }{ b-a } \\2Kc+L&=\frac { \left(K{ b }^{ 2 }+Lb+M\right)-\left(K{ a }^{ 2 }+La+M\right) }{ b-a } \\&=\frac { K(b-a)(b+a)+L(b-a) }{ b-a } \\&=K(b+a)+L\\2c&=b+a\\\\\Rightarrow c&=\frac { b+a }{ 2 }. \ _\square\end{align}\]

Does there exist a function \(f(x)\) such that \(f(0)=-1\), \(f(2)=4\) and \(f'(x)\leq 2\) for all \(x?\)

If such a function exists, then from the mean value theorem there is a number \(c\) such that \(0<c<2\) and

\[f'(c)=\frac{f(2)-f(0)}{2-0}=\frac{5}{2}.\]

But this is impossible because of the assumption \(f'(x)\leq 2\). Therefore, such a function does not exist. \(_\square\)

\(f(x)\) is a function that is continuous and differentiable in the domain \([7,15]\). If \(f(7) = 21\) and \(f'(x) \leq 14\) for all \(7\leq x \leq 15\), what is the maximum possible value of \(f(15)?\)

For \(f(x) = \sqrt{x-1}\), what is the value of \(x\) in the interval \([1,\ 38]\) that satisfies the mean value theorem?

## Mean Value Theorem for Integrals

The name of the mean value theorem may require a little explanation. For a function \(f(x)\) on an interval \([a, b]\), not necessarily continuous, one may define its **mean value**, or the "average", by the formula

\[ \text{Mean} (f) = \dfrac{1}{b-a} \displaystyle \int_{a}^{b} f(x) \, dx,\]

provided, of course, that the (Riemann) integral exists. The reason is simple: the integral gives the area under the curve \(y=f(x)\), and dividing by the length of the interval yields the "average" height of the function between \(a\) and \(b\). In picture, we must have parts of the curve \(y=f(x)\) go above the mean value, and parts go below. If \(f(x)\) is continuous, then it must cross the mean value at some point. That is the "original" mean value theorem.

Mean Value Theorem for IntegralsIf \(f(x) \) is continuous on \( [a, b] \), then there exists a point \( c\) between \(a\) and \(b\) such that

\[\int_{a}^{b} f(x) \, dx = f(c) (b-a) . \]

What does this have to do with the (actual) mean value theorem, other than the semblance of the indeterminate \(c\)? Well, if \(F(x)\) is an antiderivative of \(f(x\)), the fundamental theorem of calculus (Newton-Leibniz formula) makes the left-hand side equal to \(F(b) - F(a)\), so that we have

\[\frac{F(b)-F(a)}{b-a} = f(c) = F'(c).\]

This is precisely the mean value theorem, but for \(F(x)\).

However, that does not count as a real proof of the mean value theorem. First of all, the fundamental theorem of calculus actually relies on the mean value theorem in its proof. Secondly, here \(F(x)\) is indeed continuous on \([a,b]\) and differentiable on \( (a,b)\), but its derivative furthermore is required to be continuous for the Riemann integral above to exist. One can construct (though somewhat ad hoc) functions that are differentiable, but its derivative fails to be continuous at one point, in a way that makes it not (Riemann) integrable.

## Other Applications

We can also use the mean value theorem to prove certain inequalities.

Use the mean value theorem to prove that \( \ln(x+1) < x \) for \( x > 0. \)

Suppose that our function was \( f(t) = \ln(t + 1) - t \). Note that \( t \) is just a dummy variable as we will be using \( x \) in our interval of choice.Now, given that \( f(t) \) is defined, continuous, and differentiable over the interval \( [0, x] \), by the mean value theorem, we see that

\[\begin{align}f'(c) &= \frac{f(x) - f(0)}{x - 0}\ \text{ for } c \in(0, x)\\\frac{1}{c + 1} - 1 &= \frac{\ln(x+1) - x}{x}.\end{align} \]

By observation, we know that \( c + 1 > 1, \) so it follows that \( \frac{1}{c + 1} < 1 \).Thus, \( \frac{1}{c + 1} - 1 < 0 \).

This tells us that \( \frac{\ln(x+1) - x}{x} < 0 \implies \ln(x+1) < x\) for \(x > 0.\ _\square\)

## See Also

- Extreme Value Theorem
- Intermediate Value Theorem
- Rolle's Theorem
- Fundamental Theorem of Calculus
- Taylor's Theorem (with Lagrange Remainder)