Question

Determine whether Rolle's Theorem can be applied to f on the closed interval

[a, b].
f(x) = −x2 + 3x, [0, 3]
Yes, Rolle's Theorem can be applied.No, because f is not continuous on the closed interval [a, b].No, because f is not differentiable in the open interval (a, b).No, because f(a) ≠ f(b).
If Rolle's Theorem can be applied, find all values of c in the open interval
(a, b)
such that
f '(c) = 0.
(Enter your answers as a comma-separated list. If Rolle's Theorem cannot be applied, enter NA.)
c =

Answer 1

Rolle's theorem is applicable if
`f(a)=f(b)` and $f$ is differentiable in $(a,b)$

since it's polynomial function, it's always continuous and differentiable..

and you can easily check that $f(0)=f(-3)=0$

so it is applicable.

now, $f'(x)=-2x+3=0 \implies x=\frac32$

there is only once value (as you can imagine, the graph will be downward parabola)

Answer 2

Yes, Rolle's theorem can be applied

There is only one value of c such that f'(c) = 0, and this is c = 1.5 (or 3/2 in fraction form)

Explanation:

Yes, Rolle's theorem can be applied on this function because the function is continuous in the closed interval (it is a polynomial function) and differentiable in the open interval, and f(a) = f(b) given that:

`f(0)=-0^2+3\,(0)=0\\f(3)=-3^2+3\,(3)=-9+9=0`

Then there must be a c in the open interval for which f'(c) =0

In order to find "c", we derive the function and evaluate it at "c", making the derivative equal zero, to solve for c:

`f(x)=-x^2+3\,x\\f'(x)=-2\,x+3\\f'(c)=-2\,c+3\\0=-2\,c+3\\2\,c=3\\c=(3)/(2) =1.5`

There is a unique answer for c, and that is c = 1.5