Question

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?

Answer 1

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.

Explanation:

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)`