Answer:
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)
Step-by-step 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:
[tex]f(0)=-0^2+3\,(0)=0\\f(3)=-3^2+3\,(3)=-9+9=0[/tex]
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:
[tex]f(x)=-x^2+3\,x\\f'(x)=-2\,x+3\\f'(c)=-2\,c+3\\0=-2\,c+3\\2\,c=3\\c=\frac{3}{2} =1.5[/tex]
There is a unique answer for c, and that is c = 1.5
Rolle's theorem is applicable if [tex]f(a)=f(b)[/tex] 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)
