Question

For what values of x is `f(x)= -5x^3+x^2+4x-12` concave or convex?

Answer 1

`f(x)` is convex on `(-oo,1/15)`, concave on `(1/15,+oo)`, and has a point of inflection when `x=1/15`.

Explanation:

`f(x)` is convex when `f''(x)>0`.
`f(x)` is concave when `f''(x)<0`.

Find `f''(x)`:

`f(x)=-5x^3+x^2+4x-12`

`f'(x)=-15x^2+2x+4`

`f''(x)=-30x+2`

The concavity could change when `f''(x)=0`. This is a possible point of inflection.

`f''(x)=0`

`-30x+2=0`

`x=1/15`

Analyze the sign surrounding the point `x=1/15`. You can plug in test points to determine the sign.

When `x<1/15`, `f''(x)>0`.

When `x>1/15`, `f''(x)<0`.

When `x=15`, `f''(x)=0`.

Thus, `f(x)` is convex on `(-oo,1/15)`, concave on `(1/15,+oo)`, and has a point of inflection when `x=1/15`.

Graph of `f(x)`:

graph{-5x^3+x^2+4x-12 [-22.47, 28.85, -20.56, 5.1]}

The concavity does seem to shift very close to `x=0`.