Question

How do you determine whether the function `f(x)=2x^3-3x^2-12x+1` is concave up or concave down and its intervals?

Answer 1

Investigate the sign of the second derivative.
`f` is concave down on `(-oo, 1/2)` and concave up on `(1/2, oo)`.
`(1/2, -11/2)` is the inflection point for the graph of `f`.

Explanation:

`f(x)=2x^3-3x^2-12x+1`

`f'(x)=6x^2-6x-12`

`f''(x)= 12x - 6`

`f''(x)` is never undefined and is `0` only at `x=1/2`

So the only `x` value at which the concavity might change is `1/2`

For `x < 1/2` we find that `f''(x) < 0` and for `x > 1/2`, `f''(x) > 0`

So the graph of `f` is concave down on `(-oo, 1/2)` and concave up on `(1/2, oo)`.

`f(1/2) = 2(1/2)^3-3(1/2)^2-12(1/2)+1 = -11/2`

`(1/2, -11/2)` is the only inflection point for the graph of `f`.