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Is $f (x) = \frac{1}{x^{3}} - x^{3}$ $f(x)=1/x^3-x^3$ concave or convex at $x = 1$ $x=1$ ?

1 Answer

Jan 6, 2016

Concave up.

Explanation:

First of all get f '' (x). f '(x)= $- \frac{3}{x^{4}} - 3 x^{2}$ $-3/x^4 -3x^2$ and f ''(x) = $\frac{12}{x^{5}} - 6 x$ $12/x^5 -6x$ .

At x=1, f ''(x)=6 which f ''(x) >0. That means f(x) is concave up.

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