Given $f(x) = \frac{x^2}{x-1}$, I need to find $f(2x-3)$ I'm not clear how to proceed:
Given $f(x) = \dfrac{x^2}{x-1}$, I need to determine what $f(2x-3)$ is equal to.

Given $f(x) = \dfrac{x^2}{x-1}$, we want to find $f(2x-3)$.
To do so, we substitute $2x-3$ every place we see an x in $f(x),$ and simplify:
$$f(2x-3) = \dfrac{(2x -3)^2}{(2x-3)-1}= \dfrac{4x^2 - 12x + 9}{2x-4}$$
