My Math Forum critical points of (2-4x^2)/(1+x^2)^2

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 November 24th, 2008, 12:45 AM #1 Newbie   Joined: Nov 2008 Posts: 16 Thanks: 0 critical points of (2-4x^2)/(1+x^2)^2 how could i find the critical points of (2-4x^2)/(1+x^2)^2 i use the quadratic formula and i got +or- 1i. does that mean there are no critical points?
 November 24th, 2008, 02:39 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,879 Thanks: 1087 Math Focus: Elementary mathematics and beyond Re: critical points of (2-4x^2)/(1+x^2)^2 Since the numerator and denominator are polynomials and the denominator is never equal to 0 the function is continuous, so no critical points there. Critical points also occur where the derivative is equal to zero. $f(x)=\frac{2-4x^2}{(1+x^2)^2}$ $f'(x)=\frac{-8x(1+x^2)^2-(2-4x^2)4x(1+x^2)}{(1+x^2)^4}$ $f'(x)=\frac{8x^3-16x}{(1+x^2)^3}$ One critical point is x = 0. To find the others use 8x^3 - 16x = 0 and solve for x.
 November 24th, 2008, 03:48 AM #3 Newbie   Joined: Nov 2008 Posts: 16 Thanks: 0 Re: critical points of (2-4x^2)/(1+x^2)^2 how do you know 0 is a critical value?
 November 24th, 2008, 04:24 AM #4 Member   Joined: Feb 2008 Posts: 44 Thanks: 0 Re: critical points of (2-4x^2)/(1+x^2)^2 The derivative of this function will tell you the slope of the function for any given point right? If the derivative of the function is equal to zero, then the function itself must have zero slope, which means that the function has a maximum or minimum at that point (what you call a critical value). Now, because the derivative is a rational function, we need to solve the numerator for 0 (the only way for a fraction to equal zero is if the numerator is equal to zero). $8x^3-16x$ (the numerator) factors to $8 x \left(x^2-2\right)$ Set each factor equal to zero and solve.

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