Question

How do you find what x-values are the graph on `f(x)=(2x-3) / (x^2)` concave up and concave down?

Answer 1

The graph of this function is concave up when `x>9/2` and concave down when `x<9/2` (the function is undefined at `x=0`).

Explanation:

By the Quotient Rule, the first derivative is `f'(x)=\frac{x^2*2-(2x-3)*2x}{x^4}=\frac{2x-4x+6}{x^3}=(6-2x)/(x^3)`.

Using the Quotient Rule again gives the second derivative

`f''(x)=\frac{x^3*(-2)-(6-2x)*3x^2}{x^6}`

`=\frac{-2x-18+6x}{x^4}=(4x-18)/(x^4)`

Clearly `f''(x)>0` when `x>9/2` and `f''(x)<0` when `x<9/2`, making the graph of `f` concave up when `x>9/2` and concave down when `x<9/2` (and `x!=0`).