Question

How do you determine whether the function `f(x)=(2x-3) / (x^2)` is concave up or concave down and its intervals?

Answer 1

`f(x)=(2x-3)/(x^2)` is concave up when `x > 9/2` and concave down when `x < 9/2` (and `x!=0`).

Explanation:

By the Quotient Rule, for `x!=0`, the first derivative is `f'(x)=(x^2*2-(2x-3)*2x)/(x^4)=(-2x^2+6x)/(x^4)=(-2x+6)/(x^3)`

and the second derivative is
`f''(x)=(x^3*(-2)-(-2x+6)*3x^2)/(x^6)=(4x^3-18x^2)/(x^6)=(4x-18)/(x^4)`

Since `x^4\geq 0` for all `x`, the sign of `f''(x)` is the same as the sign of its numerator `4x-18`. This expression is positive when `4x>18\Leftrightarrow x > 9/2` and negative when `4x < 18 \Leftrightarrow x < 9/2`.

The value `x=9/2` is the first coordinate of the unique inflection point of the graph of `f`. The second coordinate is `f(9/2) = (9-3)/((9/2)^2)=6/(81/4)=24/81=8/27`