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