Is $f (x) = - x^{3} - 12 x^{2} - 14 x - 2$ $f(x)=-x^3-12x^2-14x-2$ concave or convex at $x = 0$ $x=0$ ?

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Answer 1 · 2016-02-09T20:01:27

Concave at $x = 0$ $x = 0$

The test for convexity/concavity is to check the value of the 2nd derivative $f''(x)$

• If $f''(0) > 0$ then convex at $x = 0$

• If $f''(0) < 0$ then concave at $x = 0$

Find the 2nd derivative :

$f(x) = - x^3 -12x^2 - 14x -2$

$f'(x) = -3x^2 -24x - 14$

$f''(x) = 6x - 24$

Now $f''(0) = 0 - 24 = -24 < 0$

Hence $f(x)$ is concave at $x = 0$ .

Is f(x)=-x^3-12x^2-14x-2f(x)=−x3−12x2−14x−2f(x)=-x^3-12x^2-14x-2 concave or convex at x=0x=0x=0?