##
The Mean-Value Theorem

The **Mean Value Theorem** is one of the most important
theoretical tools in Calculus. It states that if *f*(*x*) is
defined and continuous on the interval [*a*,*b*] and differentiable
on (*a*,*b*), then there is at least one number *c* in the interval
(*a*,*b*) (that is *a* < *c* < *b*) such that

The special case, when
*f*(*a*) = *f*(*b*) is known as **Rolle's
Theorem**. In this case, we have *f* '(*c*) =0. In other words,
there exists a point in the interval (*a*,*b*) which has a
horizontal tangent. In fact, the Mean Value Theorem can be stated
also in terms of slopes. Indeed, the number

is the slope of the line passing through (*a*,*f*(*a*)) and
(*b*,*f*(*b*)). So the conclusion of the Mean Value Theorem states
that there exists a point
such that the tangent line
is parallel to the line passing through (*a*,*f*(*a*)) and
(*b*,*f*(*b*)). (see Picture)

**Example.** Let
,
*a* = -1and *b*=1. We have

On the other hand, for any
,
not equal to 0, we have

So the equation

does not have a solution in *c*. This does **not**
contradict the Mean Value Theorem, since *f*(*x*) is not even
continuous on [-1,1].
**Remark.** It is clear that the derivative of a constant
function is 0. But you may wonder whether a function with
derivative zero is constant. The answer is yes. Indeed, let
*f*(*x*) be a differentiable function on an interval *I*, with
*f* '(*x*) =0, for every .
Then for any *a* and *b* in
*I*, the Mean Value Theorem implies

for some *c* between *a* and *b*. So our assumption implies

Thus
*f*(*b*) = *f*(*a*) for any *a*and *b* in *I*, which means that *f*(*x*) is constant.
**Exercise 1.** Show that the equation

2*x*^{3} + 3*x*^{2} + 6*x* + 1 = 0

has exactly one real root.

**Answer.**

**Exercise 2.** Show that

for all real numbers *a* and *b*. Try to find a more general
statement.

**Answer.**

**
**

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