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We have to find critical values of -3x^2 + 6x.
Now the critical values of a function f(x) are points where f'(x) = 0
Here f(x) = -3x^2 + 6x
f'(x) = -3*2*x + 6 = -6x + 6
Equating this to zero
=> -6x + 6 = 0
=> -6x = -6
=> x = -6 / -6
=> x = 1
Therefore the critical value of -3x^2 + 6x is at x = 1.
Let the function f(x) = -3x^2 + 6x.
We need to determine the critical values of the function y.
First, we need to determine the first derivative of the function y.
==> y = -3x^2 + 6x
==> y' = -6x + 6.
Now we know that the critical values are the derivative zeros.
==> -6x + 6 = 0
We will subtract 6 from both sides.
==> -6x = -6
==> Now we will divide by -6.
==> x = -6/-6
==> x = 1
==> Then the crtitical values of y is x = 1.
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).
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