When is this value undefined?

{ -1/2 [(1 - x^2)^(-3/2)] }  /   {1- x^2}

I've already determined that it doesn't = 0 (I set numerator = 0 and graphed it and saw it wasn't), but what value of x will make it undefined? How to solve for that? 

Update:

this is the derivative of 1/(sqrt 1 - x^2) and the domain of this original function is -1 < x< 1. 

5 Answers

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  • TomV
    Lv 7
    1 month ago
    Favourite answer

    First, the derivative of 1/√(1-x²) is:

    f(x) = 1/√(1-x²)

    f'(x) = [√(1-x²)(0) - (1)(1/2)(1-x²)^(-1/2)(-2x)]/(1-x²)

     = (1/2)(2x)/(1-x²)^(3/2)

     = x/√(1-x²)^(3/2)

     not the expression you provided.

    f'(x) = 0 at x = 0

    f'(x) is undefined if x = ±1, because that would involve a division by zero. But since -1 < x < 1, f'(x) is defined over its entire domain.

    See the plot. The function is the blue trace, the derivative is green.

    Attachment image
  • 1 month ago

    That function is defined for all real values of x between -1 and +1. It is well-defined everywhere in that interval, except at x = 0.  At x = 0, we need to use l'Hôpital's Rule to get a value 0f -0.5.

  • Pope
    Lv 7
    1 month ago

    If the numerator of a faction is zero, that does not make it undefined. You should check where the denominator is zero. That would be division by zero, which is undefined.

    Also, the expression raised to power (-3/2) cannot be negative. That would not be real.

    One more thing, your first expression is not the derivative of this:

    1/(sqrt 1 - x^2)

    Nor is it the derivative of this:

    1/sqrt(1 - x^2) 

  • fcas80
    Lv 7
    1 month ago

    Denominator (1 - x^2)^(5/3) equals zero for x = - 1 and x = + 1.  If the function is defined for - 1 < x < + 1, then the function is never undefined.

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  • rotchm
    Lv 7
    1 month ago

    In your case, When the inside of radical <0 & when denoms = 0. 

    So, (1 - x^2)^(-3/2) = √ (  (1 - x^2)^(-3) ) = 1/√( (1-x²)^3 ). Thus, when

    (1-x²)^3 < 0.

    And, when your denom 1-x² = 0. [also implied by the above]. 

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