Answer
a) $\dfrac{13}{-3x+14}$; $D=\left(-\infty,\dfrac{5}{2}\right)\cup\left(\dfrac{5}{2},\dfrac{14}{3}\right)\cup\left(\dfrac{14}{3},\infty\right)$
b) $\dfrac{6x-9}{-x+8}$; $D=\left(-\infty,2\right)\cup\left(2,8\right)\cup(8,\infty)$
c) $x$; $D=\left(-\infty,2\right)\cup(2,\infty)$
d) $\dfrac{9x-16}{-8x+33}$; $D=\left(-\infty,\dfrac{5}{2}\right)\cup\left(\dfrac{5}{2},\dfrac{33}{8}\right)\cup\left(\dfrac{33}{8},\infty\right)$
Work Step by Step
We are given the functions:
$f(x)=\dfrac{2x-1}{x-2}$
$g(x)=\dfrac{x+4}{2x-5}$
Let's note:
$D_f$=the domain of $f$
$D_g$=the domain of $g$
We have:
$D_f=(-\infty,2)\cup(2,\infty)$
$D_g=(-\infty,2.5)\cup(2.5,\infty)$
a) Determine $f\circ g$ and its domain $D_{f\circ g}$:
$(f\circ g)(x)=f(g(x))=f\left(\dfrac{x+4}{2x-5}\right)=\dfrac{2\dfrac{x+4}{2x-5}-1}{\dfrac{x+4}{2x-5}-2}$
$=\dfrac{\dfrac{2x+8-2x+5}{2x-5}}{\dfrac{x+4-4x+10}{2x-5}}=\dfrac{\dfrac{13}{2x-5}}{\dfrac{-3x+14}{2x-5}}=\dfrac{13}{-3x+14}$
As $D_g$ does not contain 2.5, we exclude 2.5 from the domain of $f\circ g$. Also because 2 is not in the domain of $f$, we must exclude the value of $x$ for which $g(x)=2$:
$\dfrac{x+4}{2x-5}=2$
$x+4=4x-10$
$3x=14$
$x=\dfrac{14}{3}$
$D_{f\circ g}=\left(-\infty,\dfrac{5}{2}\right)\cup\left(\dfrac{5}{2},\dfrac{14}{3}\right)\cup\left(\dfrac{14}{3},\infty\right)$
b) Determine $g\circ f$ and its domain $D_{g\circ f}$:
$(g\circ f)(x)=g(f(x))=g\left(\dfrac{2x-1}{x-2}\right)=\dfrac{\dfrac{2x-1}{x-2}+4}{2\dfrac{2x-1}{x-2}-5}$
$=\dfrac{\dfrac{2x-1+4x-8}{x-2}}{\dfrac{4x-2-5x+10}{x-2}}=\dfrac{\dfrac{6x-9}{x-2}}{\dfrac{-x+8}{x-2}}$
$=\dfrac{6x-9}{-x+8}$
As $D_f$ does not contain 2, we exclude 2 from the domain of $g\circ f$. Also because 2.5 is not in the domain of $g$, we must exclude the value of $x$ for which $f(x)=2.5$:
$\dfrac{2x-1}{x-2}=\dfrac{5}{2}$
$2(2x-1)=5(x-2)$
$4x-2=5x-10$
$x=8$
$D_{g\circ f}=\left(-\infty,2\right)\cup\left(2,8\right)\cup(8,\infty)$
c) Determine $f\circ f$ and its domain $D_{f\circ f}$:
$(f\circ f)(x)=f(f(x))=f\left(\dfrac{2x-1}{x-2}\right)=\dfrac{2\dfrac{2x-1}{x-2}-1}{\dfrac{2x-1}{x-2}-2}$
$=\dfrac{\dfrac{4x-2-x+2}{x-2}}{\dfrac{2x-1-2x+4}{x-2}}=\dfrac{3x}{3}=x$
As $D_f$ does not contain 2, we exclude 2 from the domain of $f\circ f$. Also because 2 is not in the domain of $f$, we must exclude the value of $x$ for which $f(x)=2$:
$\dfrac{2x-1}{x-2}=2$
$2x-1=2x-4$
$-1=-4$
No solution
$D_{f\circ f}=\left(-\infty,2\right)\cup(2,\infty)$
d) Determine $g\circ g$ and its domain $D_{g\circ g}$:
$(g\circ g)(x)=g(g(x))=g\left(\dfrac{x+4}{2x-5}\right)=\dfrac{\dfrac{x+4}{2x-5}+4}{2\dfrac{x+4}{2x-5}-5}$
$=\dfrac{\dfrac{x+4+8x-20}{2x-5}}{\dfrac{2x+8-10x+25}{2x-5}}=\dfrac{9x-16}{-8x+33}$
As $D_g$ does not contain 2.5, we exclude 2.5 from the domain of $g\circ g$. Also because 2.5 is not in the domain of $g$, we must exclude the value of $x$ for which $g(x)=2.5$:
$\dfrac{x+4}{2x-5}=\dfrac{5}{2}$
$2(x+4)=5(2x-5)$
$2x+8=10x-25$
$8x=33$
$x=\dfrac{33}{8}$
$D_{g\circ g}=\left(-\infty,\dfrac{5}{2}\right)\cup\left(\dfrac{5}{2},\dfrac{33}{8}\right)\cup\left(\dfrac{33}{8},\infty\right)$
