Question

How do you use Rolle's Theorem on a given function `f(x)`, assuming that `f(x)` is not a polynomial?

I know how to apply the theorem for polynomials, but I am unsure on how to apply it for other types of function.

Answer 1

See below.

Explanation:

To use Rolle's Theorem on any kind of function we need

a function, `f` and an interval `[a,b]`.

We determine whether the function is continuous on the closed interval `[a.b]` (We often need to explain why we think it is or is not continuous on the interval.)

We also determine whether the function is differentiable (has a derivative) on the open interval `(a,b)`. (Again, we need to be prepared to explain our answer.)

We need to determine whether `f(a) = f(b)`.

If all three conditions are met, then we may cite Rolle's Theorem to conclude that there is a `c` in `(a,b)` for which `f'(c)=0`.

The only real differences between applying Rolle's to polynomial and non-polynomial functions are

1. the function could fail to be continuous on `[a,b]` or fail to be differentiable on #(a,b),

2. it will take more to explain why we think the function does or does not satisfy the conditions for conituity and differentiabiity.

I'm not sure that this will fully answer your question, but I con't explain more without some more questions from you.

Answer 2

Here are some examples.

Explanation:

`f(x) = (-x^2)/(x^2-4)`

It is not too difficult to see that, for any `a`, we have `f(-a) = f(a)` so the third hypothesis for Rolle's is satisfied on any interval `[-a,a]`

This is a rational function, so it is continuous at every number for which it is defined.
The function is not defined, hence not continuous, at `+-2`.
(Because differentiabillity implies continuity, the function is also not differentiable at `+-2`.

For example:

`f(x) = (-x^2)/(x^2-4)` on `[-1,1]`.

`f` is continuous on `[-1,1]` because it is continuous at every `x` other than `+-2` which are not in `[-1,1]`.

`f` is differentiable on `(-1,1)` because
`f'(x) = (8x)/(x^2-4)^2` exists for all `x != +-2` (which are still not in `(-1,1)`.)

`f(-1) = f(1)` by arithmetic.

Therefore, by Rolle's Theorem, there is a `c` in `(-1,1)` with `f'(c) = 0`

Second example (same function)

`f(x) = (-x^2)/(x^2-4)` on `[-3,3]`.

`f` is not continuous on `[-3,3]` because it is not continuous at `+-2` which are in `[-3,3]`.

That is enough to tell us that we cannot use Rolle's Theorem for this function on this interval.

For the sake of the example, I will point out that `f'(x) = (8x)/(x^2-4)^2` does not exist for `x= +-2`, so the second hypothesis also fails.

We do have `f(-3) = f(3)`, so the third hypothesis is true.

NOTE THAT: although we cannot cite Rolle's Theorem on tis interval, it is still true that there is a `c` in `(-3,3)` for which `f'(c) = 0`.

`f(0) = 0` and `0` is in `(-3,3)`.

Third example (different function)

`g(x) = tanx` on `[0,pi]`.

`tan(pi/2)` does not exist, so `g` is not continuous on `[0,pi]`.
(`g` is also not differentiable at `pi/2`.)

Although `g(0)= g(pi)`, we cannot use Rolle's Theorem to conclude that there is a `c` in `(0,pi)` with `g'(c) = 0`.

In fact, since `g'(x) = sec^2 x` which is always greater then or equal to `1`, there is no `c` anywhere with `g'(c) = 0`.