Is $f (x) = \frac{{(x - 1)}^{3}}{x^{2} - 4}$ $f(x) =(x-1)^3/(x^2-4)$ concave or convex at $x = 3$ $x=3$ ?

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Answer 1 · 2017-09-24T03:25:55

concave up

$f'(x) = (3(x-1)^2)/(2x)$
$f''(x) = (6(x-1))/2$

Substitute in $x=3$
$f''(3)=(6(3-1))/2$
$=6$

When double derivatives are positive, they concave up (smiley face = positive).

Is f(x) =(x-1)^3/(x^2-4)f(x)=(x−1)3x2−4f(x) =(x-1)^3/(x^2-4) concave or convex at x=3x=3x=3?