Finding Relative Extrema for a Rational Function with a Constant

[Glissando]

Homework Statement

Find the relative extrema of the following function f(x) = (a-x)/(x²-a²)
where a is a constant, a>0

Homework Equations

Derivative of f(x), zeroes, quadratic formula

The Attempt at a Solution

I think I just screwed a small step in there because my answer doesn't work out (it's supposed to be a(1 + sqrt2) and a(1 - sqrt2)

f'(x) = [(x²+a²)(-1) - (2x)(a-x)]/(x²+a²)²

0 = -x² - a² - 2xa + 2x²

0 = x² - 2xa - a²

Quadratic formula:

x = [-b +/- sqrt(b² - 4ac)]/(2a)

x = {2xa +/- sqrt[(-2xa)² - 4(x²)(-a²)]}/(2x²)

x = [2xa +/- sqrt(4x²a² + 4x²a²)]/(2x²)

x = [2xa +/- sqrt(8x²a²)]/(2x²)

x = [2xa +/- 2sqrt(2)xa]/(2x²)

x = 2xa(1 +/- sqrt2)/(2x²)

x = a(1 +/- sqrt2)/x

): How do I get rid of the x? If you cancel it doesn't the left side become 1?

Thank you for your help! <3

 

[Dick]

It looks like you meant to say f(x)=(a-x)/(x^2+a^2) in the problem statement. So, yes, you want to solve 0=x^2-2ax-a^2. When you use the quadratic formula you put 'a'=1, 'b'=(-2a) and 'c'=(-a^2). I put quotes around the variables in the quadratic formula so as not to confuse them with the a in the problem. Notice none of them have an x in it.

 

[Glissando]

It looks like you meant to say f(x)=(a-x)/(x^2+a^2) in the problem statement. So, yes, you want to solve 0=x^2-2ax-a^2. When you use the quadratic formula you put 'a'=1, 'b'=(-2a) and 'c'=(-a^2). I put quotes around the variables in the quadratic formula so as not to confuse them with the a in the problem. Notice none of them have an x in it.

WOW that made all the difference! Thank you so much (:!