f(x) = x^2 / (x^2 - 16) f'(x) = -32x/(x^2-16)^2 x= -4,0,4 f''(x) = (96x^2 +512) / (x^2 -16)^3 how do i do this?
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Originally Posted by NeoSonata f''(x) = (96x^2 +512) / (x^2 -16)^3 how do i do this? sub the values for x you found for the stationary points. If $\displaystyle \displaystyle f''(x_0) > 0 \implies $ min and $\displaystyle \displaystyle f''(x_0) < 0 \implies $ max
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