text: # 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.
  
  We’ve answered 319,201 questions. We can answer yours, too.
explanation: The document contains a math question and a detailed answer to that question. The question asks about finding the local extremes of a specific function, and the answer provides a step-by-step solution using the first derivative test.
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distilabel_metadata.content: # 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.
  
  We’ve answered 319,201 questions. We can answer yours, too.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question and a detailed answer to that question. The question asks about finding the local extremes of a specific function, and the answer provides a step-by-step solution using the first derivative test.
  
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