# How do you find the extreme points of the function f(x) = 0.25x4 + 3x3 – 18x2 + 10 and classify them?

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The function given is f(x) = 0.25x^4 + 3x^3 – 18x^2+ 10

At the extreme points f'(x) = 0

f'(x) = 4*0.25x^3 + 9x^2 - 36x

x^3 + 9x^x - 36 = 0

=> x(x^2 + 9x - 36) = 0

=> x(x^2 + 12x - 3x - 36) = 0

=> x(x(x+ 12) - 3(x + 12)) = 0

=> x(x + 12 )(x - 3) = 0

The extreme points are at x = 0, x = 3 and x = -12

At x = 0, f(x) = 10

At x = 3, f(x) = -203/4

At x = -12, f(x) = 7786

**The extreme points are (3, -203/4) , (0 , 10) and (-12 , 7786)**

To determine the extreme points of the function, we need to determine the critical values first. The critical values are the roots of the 1st derivative of the function.

For this reason, we'll calculate the first derivative of the function:

f'(x) = 4*(1/4)*x^3 + 9x^2 - 36x

We'll cancel f'(x):

f'(x) = 0

x^3 + 9x^2 - 36x = 0

We'll factorize by x:

x*(x^2 + 9x - 36) = 0

We'll cancel each factor:

x = 0

x^2 + 9x - 36 = 0

We'll apply quadratic formula:

x1 = [-9+sqrt(81 + 144)]/2

x1 = (-9+15)/2

x1 = 3

x2 = -12

The critical points of the function are: x = 0 , x = 3 and x = -12.

To find extreme points, we'll have to determine the y coordinates for the critical values:

f(0) = 10

f(3) = 81/4 + 81 - 162 + 10

f(3) = -71 + 81/4

f(3) = -203/4

f(-12) = 5184 + 5184 - 2592 + 10

f(-12) = 7786

**The extreme values are: (0 , 10) ; (3 , -203/4) ; (-12 , 7786).**