Answer
a) $f(0)=-\frac{1}{5}$
(b) $f(1)=-\frac{3}{2}$
(c)$f(-1)=\frac{1}{8}$
(d) $f(-x)=\frac{2x-1}{3x+5}$
(e) $-f(x) \frac{-2x-1}{3x-5}$
(f) $f(x+1)=\frac{2x+3}{3x-2}$
(g) $f(2x)=\frac{4x+1}{6x-5}$
(h) $f(x+h)=\frac{2x+2h+1}{3x+3h-5}$
Work Step by Step
Given $f(x)=\frac{2x+1}{3x-5}$
a) To Evaluate $f(0)$ in the given function, substitute $0$ in the place of $x$.
$f(0)=\frac{2(0)+1}{3(0)-5}=\frac{0+1}{0-5}=-\frac{1}{5}$
(b). To Evaluate $f(1)$ in the given function, substitute $1$ in the place of $x$.
$f(1)=\frac{2(1)+1}{3(1)-5}=\frac{2+1}{3-5}=\frac{3}{-2}=-\frac{3}{2}$
(c). To Evaluate $f(−1)$ in the given function, substitute $−1$ in the place of $x$.
$f(-1)=\frac{2(-1)+1}{3(-1)-5}=\frac{-2+1}{-3-5}=\frac{-1}{-8}=\frac{1}{8}$
(d) To Evaluate $f(−x)$ in the given function, substitute $−x$ in the place of $x$.
$f(-x)=\frac{2(-x)+1}{3(-x)-5}=\frac{-2x+1}{-3x-5}=\frac{2x-1}{3x+5}$
(e) To Evaluate $−f(x)$ in the given function,
$-f(x)=-(\frac{2x+1}{3x-5}) = \frac{-2x-1}{3x-5}$
(f) To Evaluate $f(x+1)$ in the given function, substitute $x+1$ in the place of $x$.
$f(x+1)=\frac{2(x+1)+1}{3(x+1)-5}=\frac{2x+2+1}{3x+3-5}=\frac{2x+3}{3x-2}$
(g) To Evaluate $f(2x)$ in the given function, substitute $2x$ in the place of $x$.
$f(2x)=\frac{2(2x)+1}{3(2x)-5}=\frac{4x+1}{6x-5}$
(h) To Evaluate $f(x+h)$ in the given function, substitute $x+h$ in the place of $x$.
$f(x+h)=\frac{2(x+h)+1}{3(x+h)-5}=\frac{2x+2h+1}{3x+3h-5}$
