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

Asked on by elion

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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)

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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).

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