Question

If $f\left(x\right)=\ln x,g\left(x\right)=e^x$, and $h\left(x\right)=x^2$, find:

(a)role="math" localid="1647337775231" $\left(f\circ g\right)\left(x\right)$. What is the domain of $f\circ g$?

(b) $\left(g\circ f\right)\left(x\right)$. What is the domain of $g\circ f$ ?

(c) $(f\circ g)\left(5\right)$

(d)$(f\circ h)\left(x\right)$. What is the domain of $f\circ h$?

(e) $\left(f\circ h\right)\left(e\right)$

Short answer

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

The domain of $\left(f\circ g\right)$is $\left(-\infty ,\infty \right)$

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

The domain of $\left(g\circ f\right)$is $\left(-\infty ,\infty \right)$

(c) $\left(f\circ g\right)\left(5\right)=5$

(d) $\left(f\circ h\right)\left(x\right)=\ln x^2$

Domain of $\left(f\circ h\right)$is $\left(-\infty ,0\right)\cup \left(0,\infty \right)$

(e)$\left(f\circ h\right)\left(e\right)=2$

Step-by-step solution

Step 1. Given information

The Given functions are

$$\begin{gathered} f\left(x\right)=\ln x \\ g\left(x\right)=e^x \\ h\left(x\right)=x^2 \end{gathered}$$

Step 2. Part (a)

Composition of function $\left(f\circ g\right)\left(x\right)$

$$\begin{gathered} \left(f\circ g\right)\left(x\right)=f\left(g\right(x\left)\right) \\ \left(f\circ g\right)\left(x\right)=\ln \left(g\left(x\right)\right) \\ \left(f\circ g\right)\left(x\right)=\ln \left(e^x\right) \\ \left(f\circ g\right)\left(x\right)=x \end{gathered}$$

So the domain of$f\circ g$is$\left(-\infty ,\infty \right)$

Step 3. Part (b)

Composition of function $\left(g\circ f\right)\left(x\right)$

$$\begin{gathered} \left(g\circ f\right)\left(x\right)=g\left(f\right(x\left)\right) \\ \left(g\circ f\right)\left(x\right)=e^{f\left(x\right)} \\ \left(g\circ f\right)\left(x\right)=e^{\ln x} \\ \left(g\circ f\right)\left(x\right)=x \end{gathered}$$

So the domain of$g\circ f$is$\left(-\infty ,\infty \right)$

Step 4. Part (c)

Substitute $x=5$in $\left(f\circ g\right)\left(x\right)$

$$\begin{gathered} \left(f\circ g\right)\left(x\right)=x \\ \left(f\circ g\right)\left(5\right)=5 \end{gathered}$$

Step 5. Part (d)

Composition of function $\left(f\circ h\right)\left(x\right)$

$$\begin{gathered} \left(f\circ h\right)\left(x\right)=f\left(h\right(x\left)\right) \\ \left(f\circ h\right)\left(x\right)=f\left(x^2\right) \\ \left(f\circ h\right)\left(x\right)=\ln x^2 \end{gathered}$$

So the domain of$\left(f\circ h\right)$is$\left(-\infty ,0\right)\cup \left(0,\infty \right)$

Step 6. Part (e)

Substitute $x=e$in role="math" $\left(f\circ h\right)$

$$\begin{gathered} \left(f\circ h\right)\left(x\right)=\ln x^2 \\ \left(f\circ h\right)\left(e\right)=\ln e^2 \\ \left(f\circ h\right)\left(e\right)=2 \end{gathered}$$