# Local extremes of the function.Find the local extremes of the function f(x)=x^3/(x^2-1) ?

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To find out the local extremes of a function, we must do the first derivative test.

We'll determine the derivative of the function. Since the function is a fraction, we'll apply the quotient rule:

(u/v)' = (u'*v - u*v')/v^2

We'll put u = x^3

u' = 3x^2

v = x^2 - 1

v' = 2x

Now, we'll determine f'(x):

f'(x) = [3x^2(x^2 - 1) - x^3*2x]/( x^2 - 1)^2

We'll factorize the numerator by x^2:

f'(x) = x^2(3x^2 - 3 - 2x^2)/( x^2 - 1)^2

We'll combine like terms and we'll get:

f'(x) = x^2(x^2 - 3)/( x^2 - 1)^2

The function has local extreme points for values of x that cancel the first derivative.

We'll put f'(x) = 0.

x^2(x^2 - 3)/( x^2 - 1)^2 = 0

x^2(x^2 - 3) = 0

x1=x2 = 0

x^2 - 3 = 0

x^2 = 3

x3 = +sqrt3

x4 = -sqrt3

The function has local extreme points for x = -sqrt3, x=0 and x = sqrt3.