Rolle’s Theorem is working with the line of a function between two points of “equal heights” or equal y-values or equal function values. For this next Theorem, the Mean Value Theorem, these two values we are working with do not have to be equal heights.

First, let's remind ourselves what the secant line is. The secant line is the straight line between two points, think of it with the saying “as the crow flies”, or as the straightest fastest path between two points.

The Mean Value Theorem states that for two points along a line of function, there will be a point along that function between those two points that matches the gradient of the secant line between those two points. Again, seeing it visually may make more sense.

We have the graph of a function, *f(x)* below. We also have the interval of the two points *a ≤ x ≤ b* along that graphed function. The secant line is marked in green.

Roughly, we can see that there are points c_{1}, c_{2}, that have matching gradients/slopes to that of the secant.

The Mean Value Theorem: If the function f(x) is continuous on the interval a ≤ x ≤ b and it is differentiable on a < x < b,

There one or more values for *c*, between *a* and *b*, such that the tangent at the value *c* is parallel (the same gradient) to the secant line between the points (a, f(a)) and (b, f(b)),

f’(c) =

f(b) - f(a)

b - a

For the following functions, calculate *c*, the value that makes f '(c) = m by using the Mean Value Theorem, where *m* is the gradient of the secant between the end points of the interval.

a.) f(x) = x^{2} + 2x - 1 for the interval [1, 3]

First, recall the Mean Value Theorem definition:

*If a function is continuous on interval [a, b] and differentiable on (a, b), there are one or more values of c, such that f '(c) = m, where m is the gradient of the secant line between a and b.*

In other words, there is a point between two points that has the same slope as the straight line between those two surrounding points.

To calculate *c* we first need to find *m*, which is the slope of the secant line between *a* and *b*.

The formula for the slope of a straight line between two points is:

m =

f(b) - f(a)

b - a

So, with f(x) = x^{2} + 2x - 1 and the interval [1, 3]:

m =

f(b) - f(a)

b - a

m =

(3^{2} + 2(3) - 1) - (1^{2} + 2(1) - 1)

3 - 1

=

(9 + 6 - 1) - (1 + 2 - 1)

2

=

14 - 2

2

= 6

So, *m* = 6

We can now figure out f '(x) to equate f '(c) = m

f '(x) = 2x + 2, f '(c) = 2c + 2 = m = 6

2c + 2 = 6, 2c = 4, c = 2

c = 2 is the point where the slope of that point is equal to the straight line between 1 and 3 on f(*x*).

b.) f(x) = 2x^{2} -6x + 1 for the interval [2, 4]

First, recall the Mean Value Theorem :

For f(x) on [a, b]

f(b) - f(a)

b - a

, where *m* is the slope of the secant between *a* and *b*.

So first find *m*

m =

f(b) - f(a)

b - a

f(x) = 2x^{2} - 6x + 1 on [2, 4]:

m =

(2(4)^{2} - 6(4) + 1) - (2(2)^{2} + 6(2) + 1)

4 - 2

m =

(9) - (-3)

2

= 6

So m = 6

Now find f'(x) so we can equate f '(c) = m

f '(x) = 4x - 6

f '(c) = 4(c) - 6 = 6, 4c = 12, c = 3

c = 3 is the value in f(x) where its slope is equal to the slope of the secant line between the endpoints of the interval.