Answer
(a) $2$
(b) $ -2$
(c) $ \dfrac{-3x}{x^2-1}$
(d) $ -\dfrac{3x}{x^2-1}$
(e) $ \dfrac{3(x-2)}{x^2-4x+3}$
(f) $ \dfrac{6x}{4x^2-1}$
Work Step by Step
Given $f(x)=\dfrac{3x}{x^2-1}$, we have:
(a) $f(2)=\dfrac{3(2)}{(2)^2-1}=2$
(b) $f(-2)=\dfrac{3(-2)}{(-2)^2-1}=-2$
(c) $f(-x)=\dfrac{3(-x)}{(-x)^2-1}=\dfrac{-3x}{x^2-1}$
(d) $-f(x)=-\dfrac{3x}{x^2-1}$
(e) $f(x-2)=\dfrac{3(x-2)}{(x-2)^2-1}=\dfrac{3(x-2)}{x^2-4x+3}$
(f) $f(2x)=\dfrac{3(2x)}{(2x)^2-1}=\dfrac{6x}{4x^2-1}$
