What are the local extremes of the function f(x)=x^3/(x^2-1) ?

To determine the local extremes of a function, we'll have
to 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.

f(-sqrt3) =
(-sqrt3)^3/[(-sqrt3)^2 -
1]

f(-sqrt3) =
-3sqrt3/2

f(0) =
0

f(sqrt3) =
3sqrt3/2