The Substitution Rule

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 5.5 – The Substitution Rule

Composition of Functions

Definition: Composition of Functions

Let $f(x)$$f(x)$ and $g(x)$$g(x)$ be functions where the range of $g(x)$$g(x)$ is a subset of the domain of $f(x)$$f(x)$, then ${\displaystyle f(g(x))}$$\displaystyle f(g(x))$ is called the composition of $f$$f$ with $g$$g$.

NOTE: ${\displaystyle f(g(x))\ne f(x)g(x)}$$\displaystyle f(g(x))\ne f(x)g(x)$

NOTE: ${\displaystyle f(g(x))}$$\displaystyle f(g(x))$ is not necessarily the same as $g(f(x))$$g(f(x))$.

Procedure: Decomposing a Function

Decomposing a function consists of identifying:

• all the functions in the composition;
• the order of the composition, i.e., the order of the functions applied to $x$$x$.

Example 01: Decompose the function: $\sqrt{\cos (x)}$$\sqrt{\cos{(x)}}$.

Solution:

There are two (2) functions: $f(x)=\sqrt{x}$$\color{red}f(x)=\sqrt{x}$ and $g(x)=\cos (x)$$\color{green}g(x)=\cos{(x)}$. Either,

• $\sqrt{x}$$\color{red}\sqrt{x}$ replaces the $x$$x$-variable in the $\cos (x)$$\color{green}\cos{(x)}$ which results in $\cos (\sqrt{x})\ne \sqrt{\cos (x)}$${\color{green}{\cos{(}}{\color{red}\sqrt{x}}{\color{green})}}\ne\sqrt{\cos{(x)}}$; or
• $\cos (x)$$\color{green}\cos{(x)}$ replaces the $x$$x$-variable in the $\sqrt{x}$$\color{red}\sqrt{x}$ which results in $\sqrt{\cos (x)}$${\color{red}\sqrt{\color{green}{\cos{(x)}}}}$, which is the correct order. $✓$$\checkmark$ Thus, $\cos (x)$$\color{green}\cos{(x)}$ is the "innermost" function.

Therefore, $f(g(x))=\sqrt{\cos (x)}$${\color{red}f(}{\color{green}g(x)}{\color{red})}={\color{red}\sqrt{\color{green}{\cos{(x)}}}}$.

Example 02: Decompose the function: ${\displaystyle e^{\sqrt{x^2+1}}}$$\displaystyle e^\sqrt{x^2+1}$.

Solution:

There are three (3) functions: $f(x)=\sqrt{x}$$\color{red}f(x)=\sqrt{x}$, $g(x)=e^x$$\color{blue}g(x)=e^x$, and $h(x)=x^2+1$$\color{green}h(x)=x^2+1$. First,

the $x$$x$-variable in $e^x$$\color{blue}e^x$ is replaced by $\sqrt{x^2+1}$$\sqrt{x^2+1}$ implies $e^x$$\color{blue}e^x$ is the "outermost" function. Now, either

• $\sqrt{x}$$\color{red}\sqrt{x}$ replaces the $x$$x$-variable in the $x^2+1$$\color{green}x^2+1$ which results in $(\sqrt{x}){}^2+1\ne \sqrt{x^2+1}$$({\color{red}\sqrt{x}}){\color{green}^2+1}\ne\sqrt{x^2+1}$; or
• $x^2+1$$\color{green}x^2+1$ replaces the $x$$x$-variable in the $\sqrt{x}$$\color{red}\sqrt{x}$ which results in $\sqrt{x^2+1}$${\color{red}\sqrt{\color{green}{x^2+1}}}$, which is the correct order. $✓$$\checkmark$ Thus, $x^2+1$$x^2+1$ is the "innermost" function.

Therefore, ${\displaystyle g(f(h(x)))=e^{\sqrt{x^2+1}}}$$\displaystyle {\color{blue}g(}{\color{red}f(}{\color{green}h(x)}{\color{red})}{\color{blue})}={\color{blue}e}^{\color{red}\sqrt{\color{green}x^2+1}}$.

Example 03: Decompose the function: ${\displaystyle {(\tan \left(e^x\right))}^{-1}}$$\displaystyle \left(\tan{\left(e^x\right)}\right)^{-1}$.

Solution:

Since ${\displaystyle {(\tan \left(e^x\right))}^{-1}=\frac{1}{\tan \left(e^x\right)}}$$\displaystyle \left(\tan{\left(e^x\right)}\right)^{-1}=\frac{1}{\tan{\left(e^x\right)}}$, there are three (3) functions: $f(x)=\tan (x)$$\color{red}f(x)=\tan{(x)}$, $g(x)=\frac{1}{x}$$\color{blue}g(x)=\frac{1}{x}$, and $h(x)=e^x$$\color{green}h(x)=e^x$.

First, the $x$$x$-variable in $\frac{1}{x}$$\color{blue}\frac{1}{x}$ is replaced by $\tan (e^x)$$\tan{(e^x)}$ implies $\frac{1}{x}$$\color{blue}\frac{1}{x}$ is the "outermost" function. Now, either

• $\tan (x)$$\color{red}\tan{(x)}$ replaces the $x$$x$-variable in the $e^x$$\color{green}e^x$ which results in $e^{\tan (x)}\ne \tan (e^x)$${\color{green}e}^{{\color{red}\tan{(x)}}}\ne\tan{(e^x)}$; or
• $e^x$$\color{green}e^x$ replaces the $x$$x$-variable in the $\tan (x)$$\color{red}\tan{(x)}$ which results in $\tan \left(e^x\right)$$\color{red}\tan({\color{green}e^x})$, which is the correct order. $✓$$\checkmark$ Thus, $e^x$$e^x$ is the "innermost" function.

Therefore, ${\displaystyle g(f(h(x)))=\frac{1}{\tan \left(e^x\right)}}$$\displaystyle {\color{blue}g(}{\color{red}f(}{\color{green}h(x)}{\color{red})}{\color{blue})}=\color{blue}\frac{1}{\color{red}\tan({\color{green}e^x})}$.

Example 04: Decompose the function: ${\displaystyle {\cos }^{-1}(5+\ln (x))}$$\displaystyle \cos^{-1}{\!(5+\ln{(x)})}$.

Solution:

There are three (3) functions: $f(x)=5+x$$\color{red}f(x)=5+x$, $g(x)={\cos }^{-1}(x)$$\color{blue}g(x)=\cos^{-1}{\!(x)}$, and $h(x)=\ln (x)$$\color{green}h(x)=\ln{(x)}$.

NOTE: ${\cos }^{-1}(x)$$\cos^{-1}{\!(x)}$ is the arccosine function and is NOT a composition

First, the $x$$x$-variable in ${\cos }^{-1}(x)$$\color{blue}\cos^{-1}{\!(x)}$ is replaced by $5+\ln (x)$$5+\ln{(x)}$ implies ${\cos }^{-1}(x)$$\color{blue}\cos^{-1}{\!(x)}$ is the "outermost" function. Now, either

• $5+x$$\color{red}5+x$ replaces the $x$$x$-variable in the $\ln (x)$$\color{green}\ln{(x)}$ which results in $\ln \left(5+x\right)\ne 5+\ln (x)$${\color{green}\ln({\color{red}5+x})}\ne 5+\ln{(x)}$; or
• $\ln (x)$$\color{green}\ln{(x)}$ replaces the $x$$x$-variable in the $5+x$$\color{red}5+x$ which results in $5+\ln (x)$${\color{red}5+{\color{green}\ln{(x)}}}$, which is the correct order. Thus, $\ln (x)$$\ln{(x)}$ is the "innermost" function.

Thus, ${\displaystyle g(f(h(x)))={\cos }^{-1}(5+\ln \left(x\right))}$$\displaystyle {\color{blue}g(}{\color{red}f(}{\color{green}h(x)}{\color{red})}{\color{blue})}=\color{blue}\cos^{-1}({\color{red}5+\ln({\color{green}x})})$.

NOTE: Before decomposing a function, you need to identify which functions are compositions and which functions are products.

Question 01

Use your knowledge of algebra to identify which functions are compositions and which functions are products:

1. $x^2{e}^x$$x^2e^x$
2. $e^{x^2}$$e^{x^2}$
3. $\sin (\cos (x))$$\sin{(\cos{(x)})}$
4. $\cos (x)\sin (x)$$\cos{(x)}\sin{(x)}$
5. $e^{\tan (x)}$$e^{\tan{(x)}}$
6. $\ln (\cos (x))$$\ln{(\cos{(x)})}$
7. $\log (x)\sin (x)$$\log{(x)}\sin{(x)}$
8. $\sqrt{e^x}$$\sqrt{e^x}$
9. $\sin (x^2)$$\sin{(x^2)}$
10. $\sin (x{)}^2$$\sin{(x)^2}$ NOTE: The above function (#10) is equivalent to ${\sin }^2(x)$$\sin^2{\!(x)}$ and ${(\sin (x))}^2$$\left(\sin{(x)}\right)^2$. NOTE: Avoid writing $\sin x^2$$\sin x^2$ due to its ambiguity due to very poor mathematical notation.

Question 02

Use your knowledge of algebra to identify the “inside” function of the following compositions of two functions:

1. $\tan (x^2)$$\tan{(x^2)}$
2. $e^{\sqrt{x}}$$e^{\sqrt{x}}$
3. $\cos (\sin (x))$$\cos{(\sin{(x)})}$
4. $\cos (e^x)$$\cos{(e^x)}$
5. $e^{\sin (x)}$$e^{\sin{(x)}}$
6. $\ln (\tan (x))$$\ln{(\tan{(x)})}$
7. $\sin (\ln (x))$$\sin{(\ln{(x)})}$
8. $\sqrt{e^x}$$\sqrt{e^x}$
9. $e^{2x}$$e^{2x}$
10. ${\sin }^2(x)$$\sin^2{(x)}$ NOTE: Alternate ways to write the above function (#10) are $\sin {(x)}^2$$\sin{(x)}^2$ and $(\sin (x){)}^2$$(\sin{(x)})^2$.

Question 03

Use your knowledge of algebra to identify the “innermost” and “outermost” functions of the following compositions of three functions:

1. $\tan (e^{x^2})$$\tan{(e^{x^2})}$
2. ${\sin }^2(e^x)$$\sin^2{(e^x)}$
3. $e^{\sqrt{\sin (x)}}$$e^{\sqrt{\sin{(x)}}}$
4. $\sqrt[3]{\cos (\sin (x))}$$\sqrt[3]{\cos{(\sin{(x)})}}$
5. $\tan (e^{\cos (x)})$$\tan{(e^{\cos{(x)}})}$
6. $e^{\sin (x^3)}$$e^{\sin{(x^3)}}$
7. $\ln (\sqrt{\tan (x)})$$\ln{(\sqrt{\tan{(x)}})}$
8. $\sin (\ln (\cos (x)))$$\sin{(\ln{(\cos{(x)})})}$
9. $\sqrt{\cos (e^x)}$$\sqrt{\cos{(e^x)}}$
10. ${\sin }^{-1}(\cos (\ln (x)))$$\sin^{-1}{\!(\cos{(\ln{(x)})})}$

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Created: Monday, 10 January 2022 - 07:09 (EST) Modified: Tuesday, 11 January 2022 - 07:33 (EST)