## Precalculus (10th Edition)

a) $f(0)=\frac{3}{4}$ (b) $f(1)=\frac{8}{9}$ (c) $f(-1)=0$ (d) $f(-x)=1-\frac{1}{(2-x)^2}=\frac{x^2-4x+3}{x^2-4x+4}$ (e) $-f(x)=\frac{1}{(x+2)^2}-1$ (f) $f(x+1)=1-\frac{1}{(x+3)^2}$ (g) $f(2x)=1-\frac{1}{4(x+1)^2}$ (h) $f(x+h)=1-\frac{1}{(x+h+2)^2}$
Given $f(x)=1-\frac{1}{(x+2)^2}$ a) To Evaluate $f(0)$ in the given function, substitute $0$ in the place of $x$. $f(0)=1-\frac{1}{(0+2)^2}=1-\frac{1}{4}=\frac{3}{4}$ (b). To Evaluate $f(1)$ in the given function, substitute $1$ in the place of $x$. $f(1)=1-\frac{1}{(1+2)^2}=1-\frac{1}{9}=\frac{8}{9}$ (c). To Evaluate $f(−1)$ in the given function, substitute $−1$ in the place of $x$. $f(-1)=1-\frac{1}{(-1+2)^2}=1-\frac{1}{1}=1-1=0$ (d) To Evaluate $f(−x)$ in the given function, substitute $−x$ in the place of $x$. $f(-x)=1-\frac{1}{(-x+2)^2}=1-\frac{1}{(2-x)^2}=\frac{(2-x)^2-1}{(2-x)^2}=\frac{4-4x+x^2-1}{4-4x+x^2}=\frac{x^2-4x+3}{x^2-4x+4}$ (e) To Evaluate $−f(x)$ in the given function, $-f(x)=-(1-\frac{1}{(x+2)^2})=\frac{1}{(x+2)^2}-1$ (f) To Evaluate $f(x+1)$ in the given function, substitute $x+1$ in the place of $x$. $f(x+1)=1-\frac{1}{(x+1+2)^2}=1-\frac{1}{(x+3)^2}$ (g) To Evaluate $f(2x)$ in the given function, substitute $2x$ in the place of $x$. $f(2x)=1-\frac{1}{(2x+2)^2}=1-\frac{1}{4(x+1)^2}$ (h) To Evaluate $f(x+h)$ in the given function, substitute $x+h$ in the place of $x$. $f(x+h)=1-\frac{1}{(x+h+2)^2}$