Question

How do you determine if rolles theorem can be applied to `sqrt(x) - 1/5x` on the interval [0,25] and if so how do you find all the values of c in the interval for which f'(c)=0?

Answer 1

Rolle's Theorem has three hypotheses:

H1 : `f` is continuous on the closed interval `[a,b]`

H2 : `f` is differentiable on the open interval `(a,b)`.

H3 : `f(a)=f(b)`

In this question, `f(x) = sqrtx - 1/5x` , `a=0` and `b=25`.

We can apply Rolle's Theorem if all 3 hypotheses are true.
H1 : This function is continuous on its domain, `[0, oo)`, so it is continuous on `[0, 25]`

H2 : `f'(x)=1/(2sqrt(x-7))` which exists for all `x >0` (for all `x in (0,oo)`,

So `f` is differentiable on `(0, 25)`

H3 : `f(0) = 0 = f(25)`

Therefore we can apply Rolle's Theorem to `f(x) = sqrtx - 1/5x` on the interval: `[0, 25]`
.
As an algebra exercise you've also been asked to actually find the values of `c` whose existence is guaranteed by Rolle's Theorem.

To do that, find `f'(x)`, set it equal to `0` and do the algebra to solve the equation. Select any solutions in `(0, 25)`.