Given $f\left(x\right)=\frac{x+2}{x-2}andg\left(x\right)=2x+5,$Find:

(a) $f\circ g$and state its domain

(b) $(g\circ f)(-2)$

(c)$(f\circ g)(-2)$

Given: $f\left(x\right)=\frac{x+2}{x-2}andg\left(x\right)=2x+5,$

We have to find

(a) $f\circ g$and state its domain.

(b) $(g\circ f)(-2)$

(c)$(f\circ g)(-2)$

According to the definition of the composition of the function:

$(f\circ g)\left(x\right)=fg\left(x\right)---\left(1\right)$

In the argument the function $f$we include the function $g$.

From the conditions of the exercise we know that

$f\left(x\right)=\frac{x+2}{x-2}andg\left(x\right)=2x+5$

By including the given functions in expression (1) we get

$$\begin{gathered} (f\circ g)\left(x\right)=fg\left(x\right)=f(2x+5) \\ f(2x+5)=\frac{(2x+5)+2}{(2x+5)-2} \\ f(2x+5)=\frac{2x+7}{2x+3} \end{gathered}$$

Since the denominator of the fraction must not be zero then $x\ne -\frac{3}{2}$(due to the denominator of the composition of the fraction $(f\circ g)\left(x\right)=\frac{1+x}{1-x}$) then

$D_{f\circ g}=(Domainoffunctionf\circ g)=\{x:x\ne -\frac{3}{2}\}$

According to the definition of composition we have

$(g\circ f)(-2)=gf(-2)$

First we need to calculate the argument of the function g , i.e. $f(-2)$

Since the function $f\left(x\right)=\frac{x+2}{x-2}$then $f(-2)=\frac{-2+2}{-2-2}=0---\left(1\right)$

So it is now $gf(-2)=g\left(0\right)$

Now, we need to calculate the argument of the function g , i.e. g(0).

Since the function $g\left(x\right)=2x+5$then $g\left(0\right)=2\times 0+5=5---\left(2\right)$

Conclusion:$(g\circ f)(-2)=gf(-2)=g\left(0\right)=5$

$(f\circ g)(-2)$means that in the function $(f\circ g)\left(x\right)$instead of the variable x we write $-2$, i.e. $x=-2$

Since $(f\circ g)\left(x\right)=\frac{2x+7}{2x+3}---\left(1\right)$

then localid="1646499494273" $(f\circ g)(-2)=\frac{2\times (-2)+7}{2\times (-2)+3}=\frac{3}{-1}=-3$

Explanation (1): We calculated in Exercise 1(x):

$(f\circ g)\left(x\right)=fg\left(x\right)=f(2x+5)=\frac{(2x+5)+2}{(2x+5)-2}=\frac{2x+7}{2x+3}$