(a) Given that $f\left(x\right)=x^2+2andg\left(x\right)=\frac{2}{x-3}$, find f ( g( x ) ) and state its domain. What is f ( g( 5 ) ) ?

(b) If $f(x)=x+2andg(x)={\log }_{2}x$, find f ( g( x ) ) and state its domain. What is f ( g( 14 ) ?

Given functions:$f\left(x\right)=x^2+2andg\left(x\right)=\frac{2}{x-3}$

We want to

• Find f ( g( x ) ) and
• State its domain.
• What is f ( g( 5 ) )

According to the definition of the composition of the function:

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

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

From the condition of the exercise we know that

localid="1647412595673" $f\left(x\right)=x^2+2andg\left(x\right)=\frac{2}{x-3}$

By Including the given functions in expression (1) we get:

$$\begin{gathered} (f\circ g)\left(x\right)=f\left(g\right(x\left)\right)=f\left(\frac{2}{x-3}\right) \\ f\left(\frac{2}{x-3}\right)={\left(\frac{2}{x-3}\right)}^2+2 \\ f\left(\frac{2}{x-3}\right)=\frac{4}{{(x-3)}^2}+2 \end{gathered}$$

Since the denominator of the fraction must not be zero then $x\ne 3$(due to the internal function $g\left(x\right)=\frac{2}{x-3}$) and $x\ne 3$(due to the denominator of the composition of the function $(f\circ g)\left(x\right)=\frac{4}{{(x-3)}^2}+2$),

i.e.$D_{f\circ g}=\left\{x\right|x\ne 3\}=(-\infty ,3)\cup (3,\infty )$

f ( g( 5 ) ) means that in the function $f\circ g$instead of the variable x we write 5, i.e. x = 5

$$\begin{gathered} Iff\left(g\right(x\left)\right)=\frac{4}{{(x-3)}^2}+2then \\ f\left(g\right(5\left)\right)=\frac{4}{{(5-3)}^2}+2 \\ f\left(g\right(5\left)\right)=\frac{4}{{\left(2\right)}^2}+2 \\ f\left(g\right(5\left)\right)=3 \end{gathered}$$

Given functions: $f(x)=x+2andg(x)={\log }_{2}x$

We want to

• Find f ( g( x ) ) and
• State its domain.
• What is f ( g( 14 ) )

According to the definition of the composition of the function:

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

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

From the condition of the exercise we know that

$f(x)=x+2andg(x)={\log }_{2}x$

By Including the given functions in expression (1) we get:

$$\begin{gathered} (f\circ g)\left(x\right)=f\left(g\right(x\left)\right)=f\left({\log }_{2}x\right) \\ f\left({\log }_{2}x\right)={\log }_{2}x+2 \end{gathered}$$

Since the argument of the logarithmic function must be a number greater than zero then x > 0 (due to the internal function g (x) = log₂ x ) and x > 0 (due to the argument of the logarithmic function of the composition $(f\circ g)\left(x\right)={\log }_2+2$),

i.e.$D_{f\circ g}=\left\{x\right|x>0\}=(0,\infty )$

f ( g( 5 ) ) means that in the function $f\circ g$instead of the variable x we write 14, i.e. x = 14

$$\begin{gathered} Iff\left(g\right(x\left)\right)={\log }_{2}x+2then \\ f\left(g\right(14\left)\right)={\log }_{2}14+2 \end{gathered}$$