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