Q: Can you explain Rolle’s Theorem?
A: Rolle’s Theorem is a specific application of the Mean Value Theorem.
What you need to apply Rolle’s Theorem: a function that is differentiable everywhere on the interval you are looking at. Basically, you need a smooth function (rolling hills, no points, breaks, sharp corners, holes, etc…)
Rolle’s says that if the function attains the same height at two different points (and meets the “smooth function” criteria) then there must be a point in the middle somewhere where the derivative of the function is zero.
Consider the function (I just made it up for example sake):
f(x) = x2 – 8x
Now, I know:
*This is a smooth / differentiable function
*And, f(2) = -12 and f(6) = -12 [just plug in 2 and then plug in 6 to verify… 2 and 6 are numbers I picked for the sake of the example]
So, Rolle’s Thm tells me that since f(2) and f(6) both = -12 and the function is differentiable…. There is an x value somewhere between 2 and 6 where the derivative of the function is 0. Essentially, there is a “turn around point” In order to go from the point (2, -12) to the point (6, -12), you must eventually “turn around” <— which gives you a derivative of 0.
That is all Rolle’s tells you. It doesn’t tell you what the point is (and it doesn’t tell you if there is just 1 point where this happens or if there are more). Rolle’s just tells you there is at least 1 point in that interval where the derivative will be 0. You have to do the work to find that point (or points).
Let’s do it for fun:
f(x) = x2 – 8x
Take the derivative:
f ‘ (x) = 2x – 8
Set = 0 to find the turnaround point:
0 = 2x – 8
Solve for x (I won’t show those steps):
x = 4.
So, the derivative is 0 when x = 4. And of course, x = 4 is definitely on the interval from 2 to 6 as Rolle’s predicted.
***Keep in mind, this is a very basic example. It turned out the the derivative only had 1 zero anyway. The derivative could have had MANY zeroes…. Rolle’s just would have guaranteed us that one of them (or more) was on our interval!