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

Asked on by noralbbig

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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

The local extreme values of f(x)  is  f(c) , where c is the solution of f'(x) = 0.

f'(x) = 0 gives {x^3/(x^2-1)}' = 0

{ (x^3)' (x^2-1) - x^3(x^2-1)'}/(x^2-1)^2  = 0

{3x^2(x^2-1) -x^3(2x)}/(x^2-1)^2 = 0. Multiply both sides by (x^2-1)^2.

3x^2(x^2-1) -2x^4 = 0

3x^4 -3x^2 -2x^4 = 0

x^4 -3x^2 = 0

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

x^2 = 0 , x^2 = 3.

x^2 = 0 gives x = 0

x^2 = 3 gives x = sqrt3 or x = -sqrt3.

Therefore local extremes are at x = 0, x = -sqrt3 and at x = sqrt3

Therefore the local extremes are given below:

f(0) = 0.

f(-sqrt3) = (-sqrt3)^3/{(-sqrt3)^2 -1} = - 3sqrt3/(3-1) = -(3/2)sqrt3

f(sqrt3) = (3/2)sqrt.

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