## Calculus 8th Edition

This function is one-to-one. Let $x_1\neq x_2$. Then $f(x_1)=2x_1-3$ and $f(x_2)=2x_2-3$. Subtracting, we get $$f(x_1)-f(x_2)=2x_1-3-(2x_2-3)=2(x_1-x_2)$$ since $x_1\neq x_2$ then $x_1-x_2\neq 0$ so we have that $f(x_1)-f(x_2)\neq0$ i.e. $f(x_1)\neq f(x_2)$. All of this means that for every $x_1\neq x_2$ also $f(x_1)\neq f(x_2)$ so the function is one-to-one.