Is #f(x)=-3x^3+4x^2+3x-4# concave or convex at #x=-1#?

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Answer 1 · 2015-12-03T02:32:56

It is convex.

If #f''(-1)<0#, the function is concave at that point.
If #f''(-1)>0#, the function is convex at that point.

#f'(x)=-9x^2+8x+3#

#f''(x)=-18x+8#

#f''(-1)=26#

Since #26>0#, the function is convex at the point #(-1,0)#.

graph{-3x^3+4x^2+3x-4 [-9.96, 10.04, -6.12, 3.88]}

The graph does appear to be convex (forming a #uu# shape) at the specified point.