How do you determine whether the function $F(x)= 1/12X^4 + 1/6X^3-3X^2-2X+1$ is concave up or concave down and its intervals?

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Answer 1 · 2015-09-13T02:04:12

To find intervals on which the graph of $F$ is concave up and those on which it is concave down, investigate the sign of the second derivative.

For, $F(x) = 1/12x^4+1/6x^3-3x^2-2x+1$ we have

$F'(x) = 1/3x^3+1/2x^2-6x-2$ and

$F''(x) = x^2+x-6 = (x+3)(x-2)$

The only chance $F''(x)$ has to (perhaps) change sign is at $F''(x) = 0$ .

Which happens at $x=-3$ and at $x=2$

On $(-oo,-3)$ , both factor of $F''$ are negative, so $F''$ is positive and the graph of $f$ is concave up.

On $(-3,2)$ , we get $F''(x)$ is negative, so the graph is concave down.

On $(2,oo)$ , $F''(x)$ is positive, so the graph is concave up.

The graph of $F$ is concave up on $(-oo,-3)$ and on $(2,oo)$ and it is concave down on $(-3,2)$

How do you determine whether the function F(x)= 1/12X^4 + 1/6X^3-3X^2-2X+1F(x)= 1/12X^4 + 1/6X^3-3X^2-2X+1 is concave up or concave down and its intervals?