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

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### 3 Answers

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)**.