Q. 2 TB - Calculus

Question

For each function k that follows, find functions f , g, and h so that k = f ◦ g ◦ h. There may be more than one possible answer.

$$\begin{gathered} k\left(x\right)={\left(x^2+1\right)}^3 \\ k\left(x\right)=\sqrt{1+{\left(x-2\right)}^2} \\ k\left(x\right)=\sqrt{\frac{1}{3x+1}} \\ k\left(x\right)=\frac{1}{\sqrt{3x+1}} \end{gathered}$$

Step-by-Step Solution

Answer

If$k\left(x\right)={\left(x^2+1\right)}^3$and$f\left(x\right)=x^3, g\left(x\right)=\left(x+1\right), \& h\left(x\right)=x^2$ then $\left(f\circ g\circ h\right)\left(x\right)=k\left(x\right).$

If $k\left(x\right)=\sqrt{1+{\left(x-2\right)}^2}$and $f\left(x\right)=\sqrt{x}, g\left(x\right)=\left(1+x^2\right), \& h\left(x\right)=x-2$then $\left(f\circ g\circ h\right)\left(x\right)=k\left(x\right).$

If $k\left(x\right)=\sqrt{\frac{1}{3x+1}}$and $f\left(x\right)=\sqrt{x}, g\left(x\right)=\frac{1}{x+1}, \& h\left(x\right)=3x$then $\left(f\circ g\circ h\right)\left(x\right)=k\left(x\right).$

If $k\left(x\right)=\frac{1}{\sqrt{3x+1}}$and $f\left(x\right)=\frac{1}{x}, g\left(x\right)=\sqrt{x+1},\& h\left(x\right)=3x$then $\left(f\circ g\circ h\right)\left(x\right)=k\left(x\right).$

Step 1. Given information.

Given functions are the following.

$$\begin{gathered} k\left(x\right)={\left(x^2+1\right)}^3 \\ k\left(x\right)=\sqrt{1+{\left(x-2\right)}^2} \\ k\left(x\right)=\sqrt{\frac{1}{3x+1}} \\ k\left(x\right)=\frac{1}{\sqrt{3x+1}} \end{gathered}$$

Step 2. Composition of the first function.

Consider $f\left(x\right)=x^3, g\left(x\right)=\left(x+1\right), \& h\left(x\right)=x^2.$

Composition of functions.

$$\begin{gathered} f\circ g\circ h=f\left(g\right(h\left(x\right)\left)\right) \\ f\circ g\circ h==f\left(g\right(x^2\left)\right) \\ f\circ g\circ h==f(x^2+1) \\ f\circ g\circ h={(x^2+1)}^3 \\ \left(f\circ g\circ h\right)\left(x\right)=k\left(x\right) \end{gathered}$$

Step 3. Composition of the second function.

Consider $f\left(x\right)=\sqrt{x}, g\left(x\right)=\left(1+x^2\right), \& h\left(x\right)=x-2.$

Composition of functions.

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

Step 4. Composition of the third function.

Consider $f\left(x\right)=\sqrt{x}, g\left(x\right)=\frac{1}{x+1}, \& h\left(x\right)=3x.$

Composition of functions.

$$\begin{gathered} \left(f\circ g\circ h\right)\left(x\right)=f\left(g\right(h\left(x\right)\left)\right) \\ \left(f\circ g\circ h\right)\left(x\right)=f\left(g\right(3x\left)\right) \\ \left(f\circ g\circ h\right)\left(x\right)=f\left(\frac{1}{3x+1}\right) \\ \left(f\circ g\circ h\right)\left(x\right)=\sqrt{\frac{1}{3x+1}} \\ \left(f\circ g\circ h\right)\left(x\right)=k\left(x\right) \end{gathered}$$

Step 5. Composition of the fourth function.

Consider $f\left(x\right)=\frac{1}{x}, g\left(x\right)=\sqrt{x+1},\& h\left(x\right)=3x.$

Composition of functions.

$$\begin{gathered} \left(f\circ g\circ h\right)\left(x\right)=f\left(g\right(h\left(x\right)\left)\right) \\ \left(f\circ g\circ h\right)\left(x\right)=f\left(g\right(3x\left)\right) \\ \left(f\circ g\circ h\right)\left(x\right)=f\left(\sqrt{3x+1}\right) \\ \left(f\circ g\circ h\right)\left(x\right)=\frac{1}{\sqrt{3x+1}} \\ \left(f\circ g\circ h\right)\left(x\right)=k\left(x\right). \end{gathered}$$