How do you find the concavity for $f(x)=3/x^2−4$ ?

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Answer 1 · 2015-09-11T18:22:05

Investigate the sign of the second derivative.

$f(x)=3/x^2−4$

$f'(x) = (-6)/x^3$

$f''(x) = 18/x^4$

$f''$ is positive for all $x$ in the domain of $f$ , so the graph of $f$ is concave up on the domain of $f$ .

The graph of $f$ is concave up on $(-oo,0)$ and on $(0,oo)$ .

How do you find the concavity for f(x)=3/x^2−4f(x)=3/x^2−4?