# Mean Value Theorem

The **Mean Value Theorem** (MVT) or the **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 a speed of 50 mph at some point in time.

The Mean Value Theorem:Suppose that a function \(f\)

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

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

#### Contents

## 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 \(\displaystyle \frac{f(b)-f(a)}{b-a}\) 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.

## Rolle's Theorem

(See full wiki here.)

The mean value theorem is best understood by first studying a weakened version of its statement.

Rolle's Theorem:Suppose that a function \(f\) is continuous and differentiable in the interval \([a, \, b]\) and \(f(a) = f(b)\).

Then, there is a number \(c\) such that \(a<c<b\) and \(f'(c) = 0\). \(_\square\)

In other words, if a function has the same value at two points, then it must be "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 rotating the graph of the function.

## Proof

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 \((a, \, f(a))\) and \((b, \, f(b))\). This line has equation

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

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: \(h'(x)= f'(x)-\dfrac{f(b)-f(a)}{b-a} .\)

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

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

\[h'(c)=f'(c)-\dfrac{f(b)-f(a)}{b-a} =0 \Rightarrow \boxed{f'(c)=\dfrac{f(b)-f(a)}{b-a}}.\]

## 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 at any interval and that \(f'(x)=3x^2+4x+1.\) Then applying the mean value theorem gives

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

Both values of \(c\) are within the interval \([-2,2]\), so they both satisfy the condition. \(_\square\)

## 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

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

## 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. \(_\square\)

Mean Value Theorem for Integrals:Given a function \( f(x) \) on the interval [a, b], if \(h(x) \) is continuous on \( [a, b] \) and differentiable on \( (a, b) \), then there exists a point \( c\) between \(a\) and \(b\) such that \( \displaystyle \int_{a}^{b} f(x) \, dx = f(c) (b-a) \). \(_\square\)

Note:This enables you to determine the mean value of \(f(x)\) on the interval \([a, b]\). Think about it: \( f(c) \) is the average height of \(f(x)\), which enables you just to multiply \(f(c)\) by the width of the rectangle, \(b-a\), to get the area under the curve. So, by dividing both sides by \(b-a\), we can write the average value of \(f(x) \) like this: \[ \text{average value}\big |_{f(x)} = \dfrac{1}{b-a} \displaystyle \int_{a}^{b} f(x) \, dx .\]

Given the function \( f(x) = \displaystyle \int_{0}^{x} \left(t^{3} - 3t^{2} + t + 1 \right)\, dt \), find the average value of \(f(x) \) from \( x = -2\) to \(x = 4\).

We have

\[\begin{align} \text{avg}|_{f(x)} &= \displaystyle \frac{1}{4-(-2)} \int_{-2}^{4} \int_{0}^{x} \left( t^{3} - 3t^{2} + t + 1 \right)\, dt \\ &= \displaystyle \frac{1}{6} \int_{-2}^{4} \left( \frac{x^{4}}{4} - x^3 + \frac{x^2}{2} + x \right)\, dx \\ &= \displaystyle \frac{1}{6} \left[ \frac{x^{5}}{20} - \frac{x^4}{4} + \frac{x^3}{6} + \frac{x^2}{2} \right]_{-2}^{4} \\ &= \frac{9}{5}. \ _\square \end{align} \]

## See Also

**Cite as:**Mean Value Theorem.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/mean-value-theorem/