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

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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.

Is f(x)=-3x^3+4x^2+3x-4f(x)=−3x3+4x2+3x−4f(x)=-3x^3+4x^2+3x-4 concave or convex at x=-1x=−1x=-1?