Question

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

Answer 1

Concave on `(-oo,1)`; convex on `(1,+oo)`

Explanation:

The convexity and concavity of a function and determined by the sign of the second derivative.

• If `f''(a)>0`, then `f(x)` is convex at `x=a`.
• If `f''(a)<0`, then `f(x)` is concave at `x=a`.

First, find the second derivative.

`f(x)=4x^3-12x^2`
`f'(x)=12x^2-24x`
`f''(x)=24x-24`

The second derivative could change signs whenever it is equal to `0`. Find that point by setting the second derivative equal to `0`.

`24x-24=0`
`x=1`

The convexity/concavity could shift only at this point. Thus, from here, we can determine on which intervals the function will be uninterruptedly convex or concave.

Use test points around `x=1`:

When `mathbf(x<1)`:

`f''(0)=-24`

Since this is `<0`, the function is concave on the interval `(-oo,1)`.

When `mathbf(x>1)`:

`f''(2)=24`

Since this is `>0`, the function is convex on the interval `(1,+oo)`.

Always consult a graph of the original function when possible:

graph{4x^3-12x^2 [-2 5, -19.9, 5.77]}

The concavity does seem to shift around the point `x=1`. When `x<1`, the graph points downward, in the `nn` shape characteristic of concavity. When `x>1`, the graph points upward in the `uu` shape characteristic of convexity.