Rolle's Theorem

Definition - Rolle’s Theorem

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.

graph illustrating Rolle's Theorem with f(a) and f(b)

c1, c2, c3, c4, c5 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,

illustration of two points with equal y-values on a x-y graph

Example 1 - Rolle’s Theorem

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

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

a) f (x) = - x2 + 6x - 6 for 1 ≤ x ≤ 5

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

f (x) = - x2 + 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) = - x2 + 6x - 6 with 1 ≤ x ≤ 5

f (5) = - 52 + 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) = - x2 + 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) = x2 + 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) = x2 + 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) = x2 + 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.

graph showing f(x) = x^2 + 8x + 14 with interval endpoints at x = -6 and x = -2 with their f(x) values being equal and equal to 2

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) = x2 + 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