Question

Is `f(x) =-x^3-(3x-2)(x+3)` concave or convex at `x=1`?

Answer 1

Concave.

Explanation:

First, simplify `f(x)` by distributing.

`f(x)=-x^3-(3x^2+7x-6)`

`f(x)=-x^3-3x^2-7x+6`

To determine this function's concavity/convexity, we must use its second derivative:

• If `f''(1)<0`, then `f(x)` is concave at `x=1`.

• If `f''(1)>0`, then `f(x)` is convex at `x=1`.

To find `f''(x)`, use the power rule many times.

`f(x)=-x^3-3x^2-7x+6`

`f'(x)=-3x^2-6x-7`

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

Find the sign of the second derivative at `x=1`:

`f''(1)=-6(1)-6=-12`

Since this is `<0`, the function is concave at `x=1`. This means the function is sloping downwards, which resembles the `nn` shape.

We can check a graph of the original function:

graph{-x^3-(3x-2)(x+3) [-4, 2, -10, 30]}