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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.
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