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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- Jul 27th 2011, 10:48 PMNeoSonatawhere the function is concave down and concave up
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? - Jul 27th 2011, 10:55 PMpickslidesRe: where the function is concave down and concave up