Is #f(x)=-2x^5-2x^3+3x^2-x+3# concave or convex at #x=-1#?

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Answer 1 · 2016-01-29T05:09:46

Convex.

You can tell if a function is concave or convex by the sign of its second derivative:

To find the second derivative, apply the power rule to each term twice.

#f(x)=-2x^5-2x^3+3x^2-x+3#

#f'(x)=-10x^4-6x^2+6x-1#

#f''(x)=-40x^3-12x+6#

Find the sign of the second derivative at #x=-1:#

#f''(-1)=-40(-1)^3-12(-1)+6#

This mostly becomes a test of keeping track of your positives and negatives.

#f''(-1)=-40(-1)+12+6=40+18=58#

Since this is #>0#, the function is convex at #x=-1#. Convexity on a graph is characterized by a #uu# shape.

We can check the graph of the original function:

graph{-2x^5-2x^3+3x^2-x+3 [-2.5, 2, -30, 30]}