What is the critical value of the function -3x^2 + 6x ?

The critical value of the function is the value of x that
cancels the first derivative of the function.

f'(x)
= (-3x^2 + 6x)'

f'(x) = -3*2x^1
+ 6

f'(x) = -6x + 6

Now,
we'll determine the roots of the equation f'(x) = 0.

-6x +
6 = 0

We'll subtract 6 both
sides:

-6x = -6

We'll divide
by -6:

x = 1

The
critical value for f(x) is x = 1.

Now,
we'll determine the local extreme of the given function, substituting x by the critical
value.

f(1) = -3*1^2 +
6*1

f(1) = -3 +
6

f(1) =
3

The maximum point of the
function is the vertex of the parabola and it has the coordinates: (1 ;
3).