Answer
(a) $ 0$
(b) $ 0$
(c) $ \dfrac{x^2-4}{x^2}$
(d) $ -\dfrac{x^2-4}{x^2}$
(e) $ \dfrac{x^2-4x}{(x-2)^2}$
(f) $ \dfrac{x^2-1}{x^2}$
Work Step by Step
Given $f(x)=\dfrac{x^2-4}{x^2}$, we have:
(a) $f(2)=\dfrac{(2)^2-4}{(2)^2}=0$
(b) $f(-2)=\dfrac{(-2)^2-4}{(-2)^2}=0$
(c) $f(-x)=\dfrac{(-x)^2-4}{(-x)^2}=\dfrac{x^2-4}{x^2}$
(d) $-f(x)=-\dfrac{x^2-4}{x^2}$
(e) $f(x-2)=\dfrac{(x-2)^2-4}{(x-2)^2}=\dfrac{x^2-4x+4-4}{(x-2)^2}=\dfrac{x^2-4x}{(x-2)^2}$
(f) $f(2x)=\dfrac{(2x)^2-4}{(2x)^2}=\dfrac{4x^2-4}{4x^2}=\dfrac{4(x^2-1)}{4x^2}=\dfrac{x^2-1}{x^2}$
