If we have a function f(x), and there are two *x* values *a* and *b*, and we have that,

f(a) = f(b)

And f(x) is continuous for a ≤ x ≤ b and it is differentiable for a < x < b,

Then there exists a number, *c*, that is between *a* and *b* that satisfies f’(c) = 0

In other words, If there are two values (*a* and *b*) that both give the same value when inputted into the function f(x), there will be a number in the function between those two values that has a gradient of 0.

c_{1}, c_{2}, c_{3}, c_{4}, c_{5} are points between *a* and *b* where a function that has f(a) = f(b) has a point * c* in that function where the gradient on that point is equal to zero. i.e. f '(c) = 0.

Another way of looking at it is, try drawing a smooth line between two points a and b, that have f(a) = f(b), **without** having a point on that connecting line that has a gradient of zero. It is not possible!

Try drawing a smooth line between the two points below,

Show that the functions below satisfy Rolle's Theorem by:

- stating that it is continuous,
- showing that the two endpoints in the interval are equal when put into the function,
- and finding the value, c, that exists such that f '(c) = 0.

a) f (x) = - x^{2} + 6x - 6 for 1 ≤ x ≤ 5

i.) First we have to state whether or not the function on the interval is continuous.

f (x) = - x^{2} + 6x - 6 for 1 ≤ x ≤ 5 is a polynomial and is therefore continuous on 1 ≤ x ≤ 5 and differentiable on 1 < x < 5.

Note: For i.), or for any time you are asked about continuity - all polynomials are continuous and therefore differentiable so, simply stating that a function is a polynomial, shows that it is continuous and differentiable.

ii.) Now we show that the two endpoints are equal in the function.

f (x) = - x^{2} + 6x - 6 with 1 ≤ x ≤ 5

f (5) = - 5^{2} + 6(5) = -25 + 30 -6 = -1

f (1) = - (1)^{2} + 6(1) - 6 = -1 + 6 - 6 = -1

So f(1) = f(5) = -1 and this satisfies Rolle's Theorem.

iii.) Now we find the value/ values of *c* that exist as f '(c) = 0

First find f '(x)

f (x) = - x^{2} + 6x - 6

f '(x) = - 2x + 6

Solve for f '(c) = 0

f '(c) = - 2c + 6 = 0

- 2c + -6

c = 3

So we have the value c = 3 in the interval 1 ≤ x ≤ 5 that has a gradient of zero on the function f(x).

b) f (x) = x^{2} + 8x + 14 for -6 ≤ x ≤ -2

So our first part of Rolle's Theorem is stating whether the function is continuous on -6 ≤ x ≤ -2 and therefore differentiable on -6 < x < 2.

f (x) = x^{2} + 8x + 14 for -6 ≤ x ≤ -2 is a polynomial and is therefore continuous on -6 ≤ x ≤ -2 and differentiable on -6 < x < 2.

ii.) Now we show that the two endpoints are equal when put into our function, f(x).

We have f (x) = x^{2} + 8x + 14 for -6 ≤ x ≤ -2

Input our first endpoint, x = -6

f (-6) = (-6)^{2} + 8(-6) + 14

= 36 - 48 + 14 = 2

Input our second endpoint, x = -2

f (-2) = (-2)^{2} + 8(-2) + 14

= 4 - 16 + 14 = 2

We have f(-6) - f(-2) = 2, so both our endpoints are equal when put into the function. Below is a reminder of what this would look like on a graph.

iii.) Now we have confirmed the function is continuous on the open interval -6 ≤ x ≤ -2 and is differentiable on -6 < x < -2 . We have also shown that the function has equal endpoints with f(-6) = f(-2) confirming it holds for Rolle's Theorem. We now know there will be one or more values of c such that f '(c) = 0 (i.e. there is a point between the endpoints in the function that has a flat gradient.)

Now we find the value/ values of *c*.

First find f '(x)

f (x) = x^{2} + 8x + 14

f '(x) = 2x + 8

Now, to find *c*, the value/ values where the gradient is flat, we equate f '(c) = 0 and solve:

f '(c) = 2c + 8 = 0

2c = -8

c = -4

The point between the endpoints that has a flat gradient is c = -2