Is $f (x) = - x e^{x} + x^{3} - x^{2}$ $f(x)=-xe^x+x^3-x^2$ concave or convex at $x = - 2$ $x=-2$ ?

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Answer 1 · 2017-01-15T12:23:32

Since $f''(-2)<0$ , the function is concave

First, you would calculate the first and the second derivatives:

$f'(x)=(-1*e^x+(-x)*e^x)+3x^2-2x$

$=(-e^x-xe^x)+3x^2-2x$

$f''(x)=(-e^x-e^x-xe^x)+6x-2$

$=-2e^x-xe^x+6x-2$

Then you would calculate

$f''(-2)=-2e^(-2)-(-2)e^(-2)+6(-2)-2$

$=cancel(-2e^-2)cancel(+2e^-2)-12-2$

$-14<0$

Since $f''(-2)<0$ , the function is concave
graph{-xe^x+x3-x^2 [-10.035, 9.965, -7.32, 2.68]}

Is f(x)=-xe^x+x^3-x^2f(x)=−xex+x3−x2f(x)=-xe^x+x^3-x^2 concave or convex at x=-2x=−2x=-2?