Is $f (x) = - 2 x^{5} - 2 x^{3} - 12 x^{2} - 6 x + 1$ $f(x)=-2x^5-2x^3-12x^2-6x+1$ concave or convex at $x = - 1$ $x=-1$ ?

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Answer 1 · 2017-01-08T22:02:27

f(x) is concave at x=1

You would get the first and the second derivative; then:

$f'(x)=-10x^4-6x^2-24x-6$

$f''(x)=-40x^3-12x-24$

$f''(1)=-40-12-24<0$

Since $f''(1)<0$ , f(x) is concave at x=1

Is f(x)=-2x^5-2x^3-12x^2-6x+1f(x)=−2x5−2x3−12x2−6x+1f(x)=-2x^5-2x^3-12x^2-6x+1 concave or convex at x=-1x=−1x=-1?