Q. 2 - Calculus

Question

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

$$\begin{gathered} \mathrm{a}) \mathrm{k}(\mathrm{x})= \ln (\sqrt{{\mathrm{x}}^2+5}\left) \\ \mathrm{b}\right) \mathrm{k}\left(\mathrm{x}\right)= {\mathrm{e}}^{\frac{1}{\sqrt{\mathrm{x}+1}}} \\ \mathrm{c}) \mathrm{k}(\mathrm{x})= {(\ln 3\mathrm{x})}^2 \\ \mathrm{d}) \mathrm{k}\left(\mathrm{x}\right)=3 \ln (5^{\mathrm{x} }+2) \end{gathered}$$

Step-by-Step Solution

Answer

a) f(x) = ln (x) , g(x) = $\sqrt{x}$, h(x) = $x^2+5$

b)  f(x) = $e^x$ ,g(x) = $\frac{1}{\sqrt{x}}$, h(x) = x+1

c) f(x) = $x^{2 }$, g(x) = ln x , h(x) = 3x.

d) f(x) = 3x , g(x) = ln x, h(x) = $5^x+2$

Step 1. Part (a)

$$\begin{gathered} \mathrm{h}\left(\mathrm{x}\right), \mathrm{g}\left(\mathrm{x}\right),\mathrm{f}\left(\mathrm{x}\right) \mathrm{Such} \mathrm{that} \mathrm{k}\left(\mathrm{x}\right)= \mathrm{h}\left(\mathrm{x}\right)\mathrm{og}\left(\mathrm{x}\right)\mathrm{of}\left(\mathrm{x}\right). \\ \mathrm{k}\left(\mathrm{x}\right)= ln\left(\sqrt{x^2+5}\right) \\ \mathrm{f}\left(\mathrm{x}\right) = ln\left(x\right). \\ \mathrm{Explanation} : \\ \mathrm{k}\left(\mathrm{x}\right)= ln\left(\sqrt{x^2+5}\right). \\ \mathrm{Through} \mathrm{simple} \mathrm{inspection} \mathrm{one} \mathrm{can} \mathrm{say} \mathrm{that} \\ h\left(x\right) =x^2 +5 \\ g\left(x\right) = \sqrt{x} \\ f\left(x\right) = ln\left(x\right) . \end{gathered}$$

Step 2 . Part (b)

$$\begin{gathered} \mathrm{h}\left(\mathrm{x}\right), \mathrm{g}\left(\mathrm{x}\right),\mathrm{f}\left(\mathrm{x}\right) \mathrm{Such} \mathrm{that} \mathrm{k}\left(\mathrm{x}\right)= \mathrm{h}\left(\mathrm{x}\right)\mathrm{og}\left(\mathrm{x}\right)\mathrm{of}\left(\mathrm{x}\right). \\ \mathrm{k}\left(\mathrm{x}\right)= e^{\frac{1}{\sqrt{x+1}}} \\ \mathrm{Answer} : \\ \mathrm{h}\left(\mathrm{x}\right)= x+1 \\ g\left(x\right) =\frac{1}{\sqrt{x}} \\ f\left(x\right) = e^x \\ \mathrm{Explaination} : \\ \mathrm{k}\left(\mathrm{x}\right)= e^{\frac{1}{\sqrt{x+1}}} \\ \mathrm{Through} \mathrm{simple} \mathrm{inspection} \mathrm{one} \mathrm{can} \mathrm{say} \mathrm{that} \\ \mathrm{h}\left(\mathrm{x}\right) =\mathrm{x}+1 \\ \mathrm{g}\left(\mathrm{x}\right) = \frac{1}{\sqrt{\mathrm{x}}} \\ \mathrm{f}\left(\mathrm{x}\right) = {\mathrm{e}}^{\mathrm{x}} . \end{gathered}$$

Step 3. Part (c) .

$$\begin{gathered} h\left(x\right), g\left(x\right),f\left(x\right) Such that k\left(x\right)= h\left(x\right)og\left(x\right)of\left(x\right). \\ k\left(x\right)= {(ln 3x)}^2 \\ Answer : \\ h\left(x\right)=3x \\ g\left(x\right)= ln x \\ f\left(x\right)={\left(x\right)}^2 \\ Explaination : \\ k\left(x\right)= {(ln 3x)}^2 \\ \mathrm{Through} \mathrm{simple} \mathrm{inspection} \mathrm{one} \mathrm{can} \mathrm{say} \mathrm{that} \\ h\left(x\right)=3x \\ g\left(x\right)= ln x \\ f\left(x\right)={\left(x\right)}^2 \end{gathered}$$

step 4. Part (4)

$$\begin{gathered} h\left(x\right), g\left(x\right),f\left(x\right) Such that k\left(x\right)= h\left(x\right)og\left(x\right)of\left(x\right). \\ k\left(x\right)= 3 ln(5^{x }+2) \\ Answer: \\ h\left(x\right)=5^x+2 \\ g\left(x\right)=ln \left(x\right) \\ f\left(x\right)=3x \\ Explaination : \\ k\left(x\right)= 3 ln(5^{x }+2) \\ \mathrm{Through} \mathrm{simple} \mathrm{inspection} \mathrm{one} \mathrm{can} \mathrm{say} \mathrm{that} \\ h\left(x\right)=5^x+2 \\ g\left(x\right)=ln \left(x\right) \\ f\left(x\right)=3x \end{gathered}$$