# Find the extreme point of f(x)=3-3x^2.

hala718 | Certified Educator

f(x) = 3 - 3x^2

First let us find the critical points where the function f(x) changes signs or directions:

To find crtitical points we will need to differentitae f(x):

==> f'(x) = - 6x

Now the critical values are the derivative's zeros:

==> - 6x = 0

==> x = 0

Then the function changes directions when x = 0

Since the sighn of x^2 is negative, then the function has a maximum values at x= 0

==> f(0 ) = 3 - 3(0^2) = 3

Then the maximum values of f(x) is : f(0) = 3

Or the funcction has a maximum values at the point ( 0, 3)

justaguide | Certified Educator

We have to find extreme points of f(x)=3-3x^2.

First we need the critical points which are the points at which the value of the slope changes signs. And as the slope of the tangents changes sign the curve changes direction.

For that we need to differentiate f(x) = 3 - 3x^2.

Now f'(x) = 0 - 6x

Equating f'(x) = 0 - 6x to 0, we get x = 0.

Therefore the critical point is at x = 0.

At x = 0, the value of f(x) = 3 - 3*0^2 = 3

So the extreme point of the slope f(x) = 3 - 3x^2 is (0,3).

giorgiana1976 | Student

f(x)=3-3x^2.

To determine the extreme values, we'll have to determine the critical values for f(x).

For this reason, we'll have to determine the first derivative:

f'(x) = -6x

The critical values of the function are the roots of the first derivative:

-6x = 0

x = 0

The given function has an extreme value for the critical value x = 0

Since the coefficient of x^2 is negative, then the function reaches the maximum values for x = 0

f(0) = 3 - 3*0^2 = 3

The maximum value of the function is f(0) = 3, so the vertex of the parabola that represents the graph of the function has the coordinates (0,3).

neela | Student

To find the extreme value of y = 3-3x^2.

x^2 >= 0 for all x.

Therefore -3x^2 < = 0 ..........(1), for all x. As mutiplying by  a negative reverses the inequality.This means -3x^2 has the maximum value  0, when x = 0

We add 3 to both sides of the equation at (1):

3-3x^2 <= 0+3.

Therefore 3-3x^2 has the maximum value of 3, when x = 0.

3-3x^2 has no minimu., as asx -->  + or- inifinity, 3-x^2 approaches -infinity.

So the extreme value of 3-3x^2 is the maximum value of 3 when x = 0.

Therefore 3-3x^2