Is $f (x) = x^{5} - 4 x^{4} + 16 x - 4$ $f(x)=x^5-4x^4+16x-4$ concave or convex at $x = - 2$ $x=-2$ ?

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Answer 1 · 2017-03-18T14:44:06

The answer is $f (x)$ $f(x)$ is concave at $x = - 2$ $x=-2$

Let's calculate the first and second derivatives

$f (x) = x^{5} - 4 x^{4} + 16 x - 4$ $f(x)=x^5-4x^4+16x-4$

$f'(x)=5x^4-16x^3+16$

$f''(x)=20x^3-48x^2=4x^2(5x-12)$

Now, we calculate, $f''(-2)$

$f''(-2)=20*(-2)^3-48*(-2)^2$

$=-160-192=-352$

As, $f''(-2)<0$ , we conclude that $f(-2)$ is concave $nn$

Normally, we calculate the concavity over an interval for example

$]-oo, 0[$

Is f(x)=x^5-4x^4+16x-4f(x)=x5−4x4+16x−4f(x)=x^5-4x^4+16x-4 concave or convex at x=-2x=−2x=-2?