Q 41. - Precalculus Enhanced with Graphing Utilities

Question

In Problems 29 – 44, for the given functions f and g, find$\left(a\right) f\circ g \left(b\right) g\circ f \left(c\right) f\circ f \left(d\right) g\circ g$  

State the domain of each composite function.

$f\left(x\right)=x^2+1; g\left(x\right)=\sqrt{x-1}$

Step-by-Step Solution

Answer

a) $f\circ g=x$and its domain is $\{x\overline{)x\ge 1\}}$.

b) $g\circ f=\left|x\right|$and its domain is the set of all real numbers.

c) $f\circ f=x^4+2x^2+2$and its domain is  the set of all real numbers.

d) $g\circ g=\sqrt{\sqrt{x-1}-1}$ and its domain is  $\left\{x\overline{)x\ge 2}\right\}$.

Step 1. Given information

The given composite function is:

$$\begin{gathered} f\left(x\right)=x^2+1 \\ g\left(x\right)=\sqrt{x-1} \end{gathered}$$

Part (a) Step 1. Find f ∘ g and its domain.

The domain of $f$ is the set of all real numbers and g is $\{x\overline{)x\ge 1\}}$.

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

Now substitute $g\left(x\right)=\sqrt{x-1}$ in the function $f\left(g\right(x\left)\right)$.

Then the function will become $f\left(\sqrt{x-1}\right)$.

Now replace x with $\sqrt{x-1}$ in $f\left(x\right)=x^2+1$,

$$\begin{gathered} f\left(\sqrt{x-1}\right)={\left(\sqrt{x-1}\right)}^2+1 \\ =x-1+1 \\ =x \end{gathered}$$

Therefore, the domain of $f\circ g$ is $\left\{x\overline{)x\ge 1}\right\}$and $(f\circ g)\left(x\right)=x$.

Part (b) Step 1. Find g ∘ f and its domain.

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

Now substitute $x^2+1$in the function $g\left(f\right(x\left)\right)$.

Then the function will become $g(x^2+1)$.

Now replace x with $x^2+1$ in $g\left(x\right)=\sqrt{x-1}$,

$$\begin{gathered} g(x^2+1)=\sqrt{(x^2+1)-1} \\ =\sqrt{x^2+1-1} \\ =\left|x\right| \end{gathered}$$

Therefore, the domain of $g\circ f$ is the set of all real numbers and $g\circ f\left(x\right)=\left|x\right|$.

Part (c) Step 1. Find f ∘ f and its domain.

c)$f\circ f\left(x\right)=f\left(f\right(x\left)\right)$

Now substitute $f\left(x\right)=x^2+1$ in the function $f\left(f\right(x\left)\right)$.

Then the function will become $f(x^2+1)$.

Now replace x with $x^2+1$ in $f\left(x\right)=x^2+1$,

$$\begin{gathered} f(x^2+1)={(x^2+1)}^2+1=x^4+2x^2+1+1 \\ =x^4+2x^2+2 \end{gathered}$$

Therefore, the domain of $f\circ f$ is the set of all real numbers and $f\circ f\left(x\right)=x^4+2x^2+2$.

Part (d) Step 1. Find g ∘ g and its domain.

$(g\circ g)\left(x\right)=g\left(g\right(x\left)\right)$

Now substitute $g\left(x\right)=\sqrt{x-1}$ in the function $g\left(g\right(x\left)\right)$.

Then the function will become $g\left(\sqrt{x-1}\right)$.

Now replace x with $\sqrt{x-1}$ in $g\left(x\right)=\sqrt{x-1}$,

$g\left(\sqrt{x-1}\right)=\sqrt{\sqrt{x-1}-1}$

Solve the inequality to find x,

$$\begin{gathered} \sqrt{x-1}-1\ge 0 \\ \sqrt{x-1}\ge 1 \\ x-1\ge 1 \\ x\ge 2 \end{gathered}$$

Therefore, the domain of $g\circ g$ is$\{x\overline{)x\ge 2\}}$ and $g\circ g\left(x\right)=\sqrt{\sqrt{x-1}-1}$.