Question

Is `f(x) =(x-3)/(3x+1)` concave or convex at `x=0`?

Answer 1

Concave

Explanation:

Find the sign of the second derivative at `x=0`.

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

To find the first derivative, use the quotient rule.

`f'(x)=((3x+1)d/dx[x-3]-(x-3)d/dx[3x+1])/(3x+1)^2`

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

`f'(x)=(3x+1-3x+9)/(3x+1)^2`

`f'(x)=10/(3x+1)^2`

We could use the quotient rule to find the second derivative, but we can also redefine `f'(x)` using negative exponents and use the chain rule:

`f'(x)=10(3x+1)^-2`

`f''(x)=10*-2(3x+1)^-3d/dx[3x+1]`

`f''(x)=-20(3x+1)^-3(3)`

`f''(x)=-60(3x+1)^-3`

`f''(x)=-60/(3x+1)^3`

The sign of the second derivative at `x=0` is

`f''(0)=-60/(3(0)+1)^3`

`f''(0)=-60/(1)^3`

`f''(0)=-60`

Since this is `<0`, the function is concave at `x=0`.

Typically, concavity resembles the `nn` shape on a graph:

graph{(x-3)/(3x+1) [-12.4, 12.92, -6.47, 6.19]}