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.
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,
Example 1 - Rolle’s Theorem
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) = - 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.
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