Question

How do we use Rolle's Theorem to find whether function `f(x)=4x^2-8x+7` has a point where `f'(x)=0` in the interval `[-1,3]`?

Answer 1

`x=1`, for details please see below.

Explanation:

Rolle's Theorem when applied the function `f(x)` must be continuous for `x` in the given range, here `[-1,3]` and `f(x)` must be differentiable for `x` in `(-1,3)`.

Here we have `a=-1` and `b=3` and as `f(x)=4x^2-8x+7`, we have `f(-1)=19` and `f(3)=19` and hence `f(a)=f(b)`.

Now according to Rolle's Theorem, if in such a function `f(a)=f(b)`, there is one `c`, where `f'(c)=0`

As `f'(x)=8x-8` and as `8x-8=0=>x=1`, we have at `x=1`, `f'(1)=0`

and our `c` is `1`.

graph{(y-4x^2+8x-7)((x+1)^2+(y-19)^2-0.01)((x-3)^2+(y-19)^2-0.01)((x-1)^2+(y-3)^2-0.015)=0 [-4, 4, -5, 23]}