Composition of Functions : Exercise (4.8) - Solutions

Key Points 

• Some pairs of functions cannot be composed. Some pairs of functions can be composed only for certain values of $x$.
  
  
• The domain of a composite function is either the same as the domain of the first function, or else lies inside it.
  
  
• The range of a composite function is either the same as the range of the second function, or else lies inside it.

1.           Functions $f$ and $g$ are given by $f(x) = 2x + 1$ and $g(x) = 3x$.

              (a)   Calculate $(g\circ f)(1)$ and $(g\circ f)(3)$.

              (b)   Find the formula of $(g\circ f)$ and check the above images. State the domainof $(g\circ f)$.

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$\begin{array}{*{20}{l}} {f(x)=2x+1,\;g(x)=3x} \\ {} \\ {(g\circ f)(1)\;=g\left( {f(1)} \right)} \\ {\;\;\;\;\;\;\;\;\;\;=g\left( {2\times 1+1} \right)} \\ {\;\;\;\;\;\;\;\;\;\;=g(3)} \\ {\;\;\;\;\;\;\;\;\;\;=3\times 3} \\ {\;\;\;\;\;\;\;\;\;\;=9} \\ {} \\ {(g\circ f)(3)\;=g\left( {f(3)} \right)} \\ {\;\;\;\;\;\;\;\;\;\;=g\left( {2\times 3+1} \right)} \\ {\;\;\;\;\;\;\;\;\;\;=g(7)} \\ {\;\;\;\;\;\;\;\;\;\;=3\times 7} \\ {\;\;\;\;\;\;\;\;\;\;=21} \\ {} \\ {(g\circ f)(x)\;=g\left( {f(x)} \right)} \\ {\;\;\;\;\;\;\;\;\;\;=g\left( {2x+1} \right)} \\ {\;\;\;\;\;\;\;\;\;\;=3\left( {2x+1} \right)} \\ {} \\ {f\;\text{is defined for all real numbers of}\;x.} \\ \begin{array}{l}\therefore \;\;\operatorname{dom}(f)=\mathbb{R}\\3\left( {2x+1} \right)\ \text{is defined for all real numbers of}\;x.\ \end{array} \\ {\therefore \;\;\operatorname{dom}(g\circ f)=\mathbb{R}} \end{array}$

2.           The functions $f$ and $g$ are given by $f(x) = x + 2$ and $g(x) = x^2$.

              (a)   Find the formulae for $(g\circ f)$,$(g\circ g)$, $(f\circ g)$, $(f\circ f)$, and their domains.

              (b)   Find $(g\circ f)(−1)$ and $(g\circ f)(2)$.

              (c)   Find $(f\circ g)(−1)$ and $(f\circ g)(2)$.

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$\begin{array}{*{20}{l}} {f(x)=x+2,\;g(x)={{x}^{2}}} \\ {} \\ {\left( {g\circ f} \right)(x)=g\left( {f(x)} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;=g\left( {x+2} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;={{{\left( {x+2} \right)}}^{2}}} \\ {} \\ {\left( {g\circ g} \right)(x)=g\left( {g(x)} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;=g\left( {{{x}^{2}}} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;={{{\left( {{{x}^{2}}} \right)}}^{2}}} \\ {\;\;\;\;\;\;\;\;\;\;\;={{x}^{4}}} \\ {} \\ {\left( {f\circ g} \right)(x)=f\left( {g(x)} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;=f\left( {{{x}^{2}}} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;={{x}^{2}}+2} \\ {} \\ {\left( {f\circ f} \right)(x)=f\left( {f(x)} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;=f\left( {x+2} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;=x+2+2} \\ {\;\;\;\;\;\;\;\;\;\;\;=x+4} \\ {} \\ {\text{Both }\;f(x)\;\text{and }g(x)\;\text{are defined for all real numbers of}\;x.} \\ {} \\ {{{{\left( {x+2} \right)}}^{2}},\ {{x}^{4}},\ {{x}^{2}}+2\ \text{and}\ x+4\ \text{are defined for all real numbers of}\;x.\ } \\ {} \\ {\operatorname{dom}\left( {g\circ f} \right)=\operatorname{dom}(f)=\mathbb{R}} \\ {\operatorname{dom}\left( {g\circ g} \right)=\operatorname{dom}(g)=\mathbb{R}} \\ {\operatorname{dom}\left( {f\circ g} \right)=\operatorname{dom}(g)=\mathbb{R}} \\ {\operatorname{dom}\left( {f\circ f} \right)=\operatorname{dom}(f)=\mathbb{R}} \end{array}$

3.           Find the formulae for composite functions $(f\circ g)$, $(g\circ f)$ and their domains in each case.

              $\begin{array}{ll} \text{(a)}\ \ f(x)=x+1, & g(x)=2 x^{2}-x+3\\\\ \text{(b)}\ \ f(x)=x^{2}-1, & g(x)=3 x+1\\\\ \text{(c)}\ \ f(x)=-x, & g(x)=x\\\\ \text{(d)}\ \ f(x)=x^{2}, & g(x)=\sqrt{x} \end{array}$

Solutions

              $\text{(a)}\ \ f(x)=x+1, \ \ \ \ \ g(x)=2 x^{2}-x+3$

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$\begin{array}{l}\left( {f\circ g} \right)(x)=f\left( {g(x)} \right)\\\\\;\;\;\;\;\;\;\;\;\;\;\ \ \ \ \ =f\left( {2{{x}^{2}}-x+3} \right)\\\\\;\;\;\;\;\;\;\;\;\;\;\ \ \ \ \ =2{{x}^{2}}-x+3+1\\\\\;\;\;\;\;\;\;\;\;\;\;\ \ \ \ \ =2{{x}^{2}}-x+4\\\\\left( {g\circ f} \right)(x)=g\left( {f(x)} \right)\\\\\;\;\;\;\;\;\;\;\;\;\;\ \ \ \ \ =g\left( {x+1} \right)\\\\\;\;\;\;\;\;\;\;\;\;\;\ \ \ \ \ =2{{\left( {x+1} \right)}^{2}}-\left( {x+1} \right)+3\\\\\;\;\;\;\;\;\;\;\;\;\;\ \ \ \ \ =2{{x}^{2}}+3x+4\\\\\text{Both }\;f(x)\;\text{and }g(x)\;\text{are defined for all real numbers of}\;x.\\\\\operatorname{dom}(f)=\mathbb{R}\;\text{and}\;\operatorname{dom}(g)=\mathbb{R}\\\\\text{Similarly, both }2{{x}^{2}}-x+4\;\text{and }2{{x}^{2}}+3x+4\;\text{are }\\\text{defined for all real numbers of}\;x.\\\\\operatorname{dom}\left( {f\circ g} \right)=\mathbb{R}\;\text{and}\;\operatorname{dom}\left( {g\circ f} \right)=\mathbb{R}\end{array}$

              $\text{(b)}\ \ f(x)=x^{2}-1, \ \ \ \ \ g(x)=3 x+1$

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$\begin{array}{*{20}{l}} {\left( {f\circ g} \right)(x)=f\left( {g(x)} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;=f\left( {3x+1} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;={{{\left( {3x+1} \right)}}^{2}}-1} \\ {\;\;\;\;\;\;\;\;\;\;\;=9{{x}^{2}}+6x} \\ {} \\ {\left( {g\circ f} \right)(x)=g\left( {f(x)} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;=g\left( {{{x}^{2}}-1} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;=3\left( {{{x}^{2}}-1} \right)+1} \\ {\;\;\;\;\;\;\;\;\;\;\;=3{{x}^{2}}-2} \\ {} \\ \begin{array}{l}\text{Both }\;f(x)\;\text{and }g(x)\;\text{are defined for all real numbers of}\;x.\\\\\therefore \ \ \operatorname{dom}(f)=\mathbb{R}\;\text{and}\;\operatorname{dom}(g)=\mathbb{R}\\\\\text{Similarly both }9{{x}^{2}}+6x\;\text{and }3{{x}^{2}}-2\;\text{are defined for }\\\text{all real numbers of}\;x.\end{array} \\ {} \\ {\therefore \ \ \operatorname{dom}\left( {f\circ g} \right)=\mathbb{R}\ \ \text{and}\ \operatorname{dom}\left( {g\circ f} \right)=\mathbb{R}} \end{array}$

              $\text{(c)}\ \ f(x)=-x, \ \ \ \ \ g(x)=x$

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$ \begin{array}{l}\left( {f\circ g} \right)(x)=f\left( {g(x)} \right)\\\\\;\;\;\;\;\;\;\;\;\;\;\ \ \ =f\left( x \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ =-x\\\\\left( {g\circ f} \right)(x)=g\left( {f(x)} \right)\\\\\;\;\;\;\;\;\;\;\;\;\;\ \ \ =g\left( {-x} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ =-x\\\\\text{Both }\;f(x)\;\text{and }g(x)\;\text{are defined for all real numbers of}\;x.\\\\\operatorname{dom}(f)=\mathbb{R}\;\text{and}\;\operatorname{dom}(g)=\mathbb{R}\\\\\text{Also}-x\ \text{is}\ \text{defined for all real numbers of}\;x.\\\\\operatorname{dom}\left( {f\circ g} \right)=\mathbb{R}\ \text{and}\ \operatorname{dom}\left( {g\circ f} \right)=\mathbb{R}\end{array}$

              $\text{(d)}\ \ f(x)=x^2, \ \ \ \ \ g(x)=\sqrt{x}$

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$\begin{array}{*{20}{l}} {\left( {f\circ g} \right)(x)=f\left( {g(x)} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;\;\;=f\left( {\sqrt{x}} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;\;\;={{{\left( {\sqrt{x}} \right)}}^{2}}} \\ {\;\;\;\;\;\;\;\;\;\;\;\;\;=x} \\ {} \\ {\left( {g\circ f} \right)(x)=g\left( {f(x)} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;\;\;=g\left( {{{x}^{2}}} \right)} \\ {\;\;\;\;\;\;\;\;\;\;\;\;\;=\sqrt{{{{x}^{2}}}}} \\ {\;\;\;\;\;\;\;\;\;\;\;\;\;=x} \\ {} \\ {f(x)\;\text{is defined for all real numbers of}\;x.} \\ {} \\ {\operatorname{dom}(f)=\mathbb{R}\;} \\ {} \\ {g(x)\;\text{is defined for }x\ge 0,\;x\in \mathbb{R}} \\ {} \\ {\operatorname{dom}(g)=\left\{ {x\;|\;x\ge 0,\;x\in \mathbb{R}} \right\}} \\ {} \\ \begin{array}{l}x\ \text{is defined for all real numbers of}\;x\ \text{but}\\\sqrt{x}\ \text{is defined for}\ x\ge 0.\ \\\\\operatorname{dom}\left( {f\circ g} \right)=\operatorname{dom}(g)=\left\{ {x\;|\;x\ge 0,\;x\in \mathbb{R}} \right\}\\\end{array} \\ {\operatorname{dom}\left( {g\circ f} \right)=\operatorname{dom}(f)=\mathbb{R}} \end{array}$

4.           A function $f$ is given by $f(x)=x+1$. Find the function $g:\mathbb{R}\to \mathbb{R}$ in each of the following:

              $\begin{array}{l} \text{(a)}\ \ \left( {g\circ f} \right)(x) = x^2+5x+5\\\\ \text{(b)}\ \ \left( {f\circ g} \right)(x) = x^2+5x+5 \end{array}$

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$\begin{array}{l} \ \ \ \ \ \ \ \ f(x)=x+1,\ \ g:\mathbb{R}\to \mathbb{R}\\\\ \text{(a)}\;\;\left( {g\circ f} \right)(x)={{x}^{2}}+5x+5\\\\ \ \ \ \ \ \ \ \ g\left( {f(x)} \right)={{x}^{2}}+5x+5\\\\ \ \ \ \ \ \ \ \ g\left( {x+1} \right)={{x}^{2}}+2x+1+3x+3+1\\\\ \ \ \ \ \ \ \ \ g\left( {x+1} \right)={{(x+1)}^{2}}+3(x+1)+1\\\\ \therefore \ \ \ g(x)={{x}^{2}}+3x+1\\\\\\ \text{(b)}\;\;\left( {f\circ g} \right)(x)={{x}^{2}}+5x+5\\\\ \ \ \ \ \ \ \ \ f\left( {g(x)} \right)={{x}^{2}}+5x+5\\\\ \ \ \ \ \ \ \ \ g(x)+1={{x}^{2}}+5x+5\\\\ \therefore \ \ \ g(x)={{x}^{2}}+5x+4 \end{array}$

5.           If $g:\mathbb{R}\to \mathbb{R}$ is given by $g(x)=x^2+3$, find the function $g$ such that

              $\begin{array}{l} \text{(a)}\ \ \left( {f\circ g} \right)(x)= 4x^2+3\\\\ \text{(b)}\ \ \left( {g\circ f} \right)(x)= 4x^2+3 \end{array}$

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$\begin{array}{l}\ \ \ \ \ \ \ g:\mathbb{R}\to \mathbb{R},\ \ \ g(x)={{x}^{2}}+3\\\\\text{(a)}\;\;\left( {f\circ g} \right)(x)=4{{x}^{2}}+3\\\\\ \ \ \ \ \ \ \ f\left( {g(x)} \right)=4{{x}^{2}}+3\\\\\ \ \ \ \ \ \ \ f\left( {{{x}^{2}}+3} \right)=4{{x}^{2}}+12-9\\\\\ \ \ \ \ \ \ \ f\left( {{{x}^{2}}+3} \right)=4\left( {{{x}^{2}}+3} \right)-9\\\\\therefore \ \ \ \ \ \ \ f\left( x \right)=4x-9\\\\\\\text{(b)}\;\;\left( {g\circ f} \right)(x)=4{{x}^{2}}+3\\\\\ \ \ \ \ \ \ \ g\left( {f(x)} \right)=4{{x}^{2}}+3\\\\\ \ \ \ \ \ \ \ {{\left( {f(x)} \right)}^{2}}+3=4{{x}^{2}}+3\\\\\ \ \ \ \ \ \ \ \ {{\left( {f(x)} \right)}^{2}}=4{{x}^{2}}\\\\\therefore \ \ \ \ \ \ \ f(x)=\pm \ 2x\end{array}$

6.           Functions $f$ and $g$ are given by $f(x)=px-2$ where $p$ is a constant and $g(x)=4x+3$. Find the value of $p$ for which $\left( {f\circ g} \right)(x) = \left( {g\circ f} \right)(x)$.

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$\begin{array}{l}\ \ \ \ f(x)=px-2,\ \ \ g(x)=4x+3\\\\\ \ \ \;\left( {f\circ g} \right)(x)=\left( {g\circ f} \right)(x)\\\\\ \ \ \ f\left( {g(x)} \right)=g\left( {f(x)} \right)\\\\\ \ \ \ f\left( {4x+3} \right)=g\left( {px-2} \right)\\\\\ \ \ \ p\left( {4x+3} \right)-2=4\left( {px-2} \right)+3\\\\\ \ \ \ 4px+3p-2=4px-8+3\\\\\ \ \ \ 3p-2=-5\\\\\ \ \ \ p=-1\end{array}$

7.           Let $f$ and $g$ be functions given by $f(x) = 3x− 1$ and $g(x) = x+7$. Find the formulae of $f^{-1}\circ g,\ g^{-1}\circ f $ and state their domains. What are the values of $\left(f^{-1}\circ g\right)(3)$ and $\left(g\circ f^{-1}\right)(2)$.

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$\begin{array}{l}\ \ \ \ f(x)=3x-1,\ \ \ g(x)=x+7\\\\\ \ \ \;\text{Let}\ {{f}^{{-1}}}(x)=y\\\\\ \ \ \ f\left( y \right)=x\\\\\ \ \ \ 3y-1=x\\\\\therefore \ \ y=\displaystyle \frac{{x+1}}{3}\\\\\therefore \ \ {{f}^{{-1}}}(x)=\displaystyle \frac{{x+1}}{3}\\\\\therefore \ \ \left( {{{f}^{{-1}}}\circ g} \right)(x)={{f}^{{-1}}}\left( {g(x)} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{f}^{{-1}}}\left( {x+7} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{x+7+1}}{3}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{x+8}}{3}\\\\\ \ \ \ \text{Both }f(x)\ \text{and }g(x)\ \text{are defined for all real numbers of }x.\\\\\therefore \ \operatorname{dom}(f)=\mathbb{R},\ \operatorname{dom}(g)=\mathbb{R}\\\\\ \ \displaystyle \frac{{x+8}}{3}\ \text{ defined for all real numbers of }x.\\\\\therefore \ \operatorname{dom}({{f}^{{-1}}}\circ g)=\mathbb{R}\\\\\ \ \ \ \text{Let}\ {{g}^{{-1}}}(x)=z\\\\\ \ \ \ g(z)=x\\\\\ \ \ \ z+7=x\\\\\ \ \ \ z=x-7\\\\\therefore \ \ {{g}^{{-1}}}(x)=x-7\\\\\therefore \ \ \left( {{{g}^{{-1}}}\circ f} \right)(x)={{g}^{{-1}}}\left( {f(x)} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{g}^{{-1}}}\left( {3x-1} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =3x-1-7\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =3x-8\\\\\ \ \ \ \text{Both }f(x)\ \text{and }g(x)\ \text{are defined for all real numbers of }x.\\\\\therefore \ \operatorname{dom}(f)=\mathbb{R},\ \operatorname{dom}(g)=\mathbb{R}\\\\\ \ 3x-8\ \text{ defined for all real numbers of }x.\\\\\therefore \ \operatorname{dom}({{g}^{{-1}}}\circ f)=\mathbb{R}\\\\\therefore \ \left( {{{f}^{{-1}}}\circ g} \right)(3)=\displaystyle \frac{{3+8}}{3}=\displaystyle \frac{{11}}{3}\\\\\ \ \ \ \ \left( {g\circ {{f}^{{-1}}}} \right)(2)=g\left( {{{f}^{{-1}}}(2)} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =g\left( {\displaystyle \frac{{2+1}}{3}} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =g\left( 1 \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1+7\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =8\end{array}$

8.           Let the functions $f$ and $g$ be given by $f(x) = 2x−1$ and $g(x) =\displaystyle \frac{2x+3}{x-1}$.

Find the composite function $f\circ g$.

Find the inverse function $f^{-1}$ and $g^{-1}$.

Evaluate $\left(f\circ g^{-1}\right)(1)$ and $\left(f^{-1}\circ g^{-1}\right)(1)$.

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$\begin{array}{l}\ f(x)=2x-1\\\ g(x)=\displaystyle \frac{{2x+3}}{{x-1}}\\\\\ \left( {f\circ g} \right)(x)=f\left( {g(x)} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =f\left( {\displaystyle \frac{{2x+3}}{{x-1}}} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2\left( {\displaystyle \frac{{2x+3}}{{x-1}}} \right)-1\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{4x+6-x+1}}{{x-1}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{3x+7}}{{x-1}}\\\\\ \text{Let}\ {{f}^{{-1}}}(x)=y,\ \text{then}\\\ f(y)=x\\\ 2y-1=x\\\ y=\displaystyle \frac{{x+1}}{2}\\\ {{f}^{{-1}}}(x)=\displaystyle \frac{{x+1}}{2}\\\\\text{Let}\ {{g}^{{-1}}}(x)=z,\ \text{then}\\\ g(z)=x\\\ \displaystyle \frac{{2z+3}}{{z-1}}=x\\\ 2z+3=xz-x\\\ xz-2z=x+3\\\ z(x-2)=x+3\\\ z=\displaystyle \frac{{x+3}}{{x-2}}\\\ {{g}^{{-1}}}(x)=\displaystyle \frac{{x+3}}{{x-2}}\\\\\left( {f\circ {{g}^{{-1}}}} \right)(1)=f\left( {{{g}^{{-1}}}(1)} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =f\left( {\displaystyle \frac{{1+3}}{{1-2}}} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =f\left( {-4} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2\left( {-4} \right)-1\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-9\\\\\left( {{{f}^{{-1}}}\circ {{g}^{{-1}}}} \right)(1)={{f}^{{-1}}}\left( {{{g}^{{-1}}}(1)} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{f}^{{-1}}}\left( {\displaystyle \frac{{1+3}}{{1-2}}} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{f}^{{-1}}}\left( {-4} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{-4+1}}{2}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\displaystyle \frac{3}{2}\end{array}$

9.           Let the functions $f$, $g$ and $h$ be 𝑓(𝑥) = $f(x) = x−2$, $g(x) = x^3$ and $h(x) = 4x$. Show that $\left(h\circ g\right)\circ f= h\circ \left(g\circ f\right)$.

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$ \begin{array}{l}\ f(x)=x-2\\\ g(x)={{x}^{3}}\\\ h(x)=4x\\\\\ \left( {h\circ g} \right)(x)=h\left( {g(x)} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =h\left( {{{x}^{3}}} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4{{x}^{3}}\\\ \\\ \left( {\left( {h\circ g} \right)\circ f} \right)(x)=\left( {h\circ g} \right)\left( {f(x)} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left( {h\circ g} \right)\left( {x-2} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4{{\left( {x-2} \right)}^{3}}\\\\\ \left( {g\circ f} \right)(x)=g\left( {f(x)} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =g\left( {x-2} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\left( {x-2} \right)}^{3}}\\\ \\\ \left( {h\circ \left( {g\circ f} \right)} \right)(x)=h\left( {\left( {g\circ f} \right)(x)} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =h\left( {{{{\left( {x-2} \right)}}^{3}}} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4{{\left( {x-2} \right)}^{3}}\\\\\ \therefore \ \ \left( {\left( {h\circ g} \right)\circ f} \right)(x)=\left( {h\circ \left( {g\circ f} \right)} \right)(x)\\\ \therefore \ \ \left( {h\circ g} \right)\circ f=h\circ \left( {g\circ f} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{array}$

Composition of Functions : Problems and Solutions

1.      Two functions are defined by $ \displaystyle f:x\mapsto ax+1$ and $ \displaystyle \ g:x\mapsto \frac{{4b}}{{x-1}},x\ne 1$ where $ \displaystyle a$ and $ \displaystyle b$ are constants. Given that $ \displaystyle f(a) = g(b)$ and $ \displaystyle f\left( {\frac{1}{a}} \right)=g\left( {\frac{1}{b}} \right)$ find the possible values of $ \displaystyle a$ and $ \displaystyle b$.

Show Solution

$ \displaystyle f(x)=ax+1\ ,g(x)=\frac{{4b}}{{x-1}},x\ne 1$

$ \displaystyle f(a)=g(b)$

$ \displaystyle \therefore {{a}^{2}}+1=\frac{{4b}}{{b-1}}$

$ \displaystyle \therefore {{a}^{2}}=\frac{{4b}}{{b-1}}-1$

$ \displaystyle f\left( {\frac{1}{a}} \right)=g\left( {\frac{1}{b}} \right)$

$ \displaystyle \therefore a\left( {\frac{1}{a}} \right)+1=\frac{{4b}}{{\frac{1}{b}-1}}$

$ \displaystyle \therefore \frac{{4{{b}^{2}}}}{{1-b}}=2\Rightarrow 2{{b}^{2}}+b-1=0$

$ \displaystyle \therefore (2b-1)(b+1)=0$

$ \displaystyle \therefore b=\frac{1}{2}\ \text{or }b=-1$

$ \displaystyle \text{When }b=\frac{1}{2},$

$ \displaystyle {{a}^{2}}=\frac{{4\left( {\frac{1}{2}} \right)}}{{\left( {\frac{1}{2}} \right)-1}}-1=-5\notin R$

$ \displaystyle \text{When }b=-1,$

$ \displaystyle {{a}^{2}}=\frac{{4\left( {-1} \right)}}{{\left( {-1} \right)-1}}-1=1$

$ \displaystyle \therefore a=\pm 1$

2.      A function $ \displaystyle f$ is defined by $ \displaystyle f:x\mapsto \frac{6}{{x-2}},x\ne 2$. Express $ \displaystyle (f\cdot f)(x)$ in the form $ \displaystyle \frac{{ax+b}}{{c-x}}$ stating the values of $ \displaystyle a, b$ and $ \displaystyle c$.

Show Solution

$\displaystyle f(x)=\frac{6}{{x-2}},x\ne 2$

$ \displaystyle (f\cdot f)(x)=f\left( {f(x)} \right)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =f\left( {\frac{6}{{x-2}}} \right)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{6}{{\frac{6}{{x-2}}-2}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{6}{{\frac{{6-2x+4}}{{x-2}}}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{6x-12}}{{10-2x}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{3x-6}}{{5-x}}$

$ \displaystyle \therefore \ \ \ \frac{{ax+b}}{{c-x}}=\frac{{3x-6}}{{5-x}}$

$ \displaystyle \therefore \ \ a=3,b=-6\ \text{and}\ c=5$

3.      A function $ \displaystyle f$ is defined by $ \displaystyle f:x\mapsto 2-\frac{1}{x},x\ne 0$. Solve the equation $ \displaystyle (f\cdot f)(x)=f(x)$.

Show Solution

$ \displaystyle \ \ \ f(x)=2-\frac{1}{x},x\ne 0$

$ \displaystyle \ \ \ \left( {f\cdot f} \right)(x)=f(x)$

$ \displaystyle \therefore f\left( {f(x)} \right)=f(x)$

$ \displaystyle \therefore f\left( {2-\frac{1}{x}} \right)=2-\frac{1}{x}$

$ \displaystyle \therefore 2-\frac{1}{{2-\frac{1}{x}}}=2-\frac{1}{x}$

$ \displaystyle \therefore \frac{x}{{2x-1}}=\frac{1}{x}$

$ \displaystyle \therefore {{x}^{2}}-2x+1=0$

$ \displaystyle \therefore {{(x-1)}^{2}}=0$

$ \displaystyle \therefore x=1$

4.      Let $ \displaystyle f:R\to R$ and $\displaystyle g:R\to R$ be $\displaystyle f(x)=px+5, g(x)=qx-3$ where $\displaystyle p\ne 0, q\ne 0$. If $ \displaystyle (g\cdot f):R\to R$ is the identity function on $\displaystyle R$, then prove that $\displaystyle p$ is the reciprocal of $\displaystyle q$. Hence find the values of $\displaystyle p$ and $\displaystyle q$.

Show Solution

$ \displaystyle f:R\to R,f(x)=px+5$

$ \displaystyle g:R\to R,g(x)=qx-3$

$ \displaystyle p\ne 0\ \text{and}\ q\ne 0$

$ \displaystyle \left( {g\cdot f} \right)(x)=I(x)$

$ \displaystyle \therefore g\left( {f(x)} \right)=x$

$ \displaystyle \therefore g\left( {px+5} \right)=x$

$ \displaystyle \therefore q\left( {px+5} \right)-3=x$

$ \displaystyle \therefore pqx+\left( {5q-3} \right)=x+0$

$ \displaystyle \therefore pq=1\Rightarrow p=\frac{1}{q}$

$ \displaystyle \ \ \ 5q-3=0\Rightarrow q=\frac{3}{5}$

$ \displaystyle \therefore p=\frac{5}{3}$

5.      If $ \displaystyle f$ and $ \displaystyle g$ are functions such that $ \displaystyle f(x) = 2x - 1$ and $ \displaystyle (g\cdot f)(x) = 4x^2 - 2x - 3$, find the formula of $ \displaystyle g$ in simplified form.

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$ \displaystyle \begin{array}{l}f(x)=2x-1\\\\\left( {g\cdot f} \right)(x)=4{{x}^{2}}-2x-3\\\\g\left( {f(x)} \right)=4{{x}^{2}}-2x-3\\\\g\left( {2x-1} \right)=4{{x}^{2}}-2x-3\\\\g\left( {2x-1} \right)=4{{x}^{2}}-4x+1+2x-1-3\\\\g\left( {2x-1} \right)={{(2x-1)}^{2}}+(2x-1)-3\\\\\therefore g(x)={{x}^{2}}+x-3\end{array}$

6.      A function f is defined by $ \displaystyle f:x\mapsto 2x^2 - 12x + 7$ for $ \displaystyle x\in R$ . Express $ \displaystyle f(x)$ in the form $ \displaystyle a(x - b)^2 - c$ and state the valueof $ \displaystyle a, b$ and $ \displaystyle c$. Another function g is defined by $ \displaystyle g:x\mapsto 2x + k$ for $ \displaystyle x\in R$. Find the value of $ \displaystyle k$ for which $ \displaystyle (g\cdot f)(x) = 0$ has two equal roots.

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$ \displaystyle \begin{array}{l}\ \ \ f(x)=2{{x}^{2}}-12x+7\\\\\therefore f(x)=2({{x}^{2}}-6x+9)-11\\\\\therefore f(x)=2{{(x-3)}^{2}}-11\\\\\ \ \ f(x)=a{{(x-b)}^{2}}-c\ \text{(given)}\\\\\therefore a{{(x-b)}^{2}}-c=2{{(x-3)}^{2}}-11\\\\\therefore a=2,\ b=3\ \text{and }c=11\\\\\ \ \ g(x)=2x+k\\\\\ \ \ \left( {g\cdot f} \right)(x)=0\\\\\therefore g\left( {f(x)} \right)=0\\\\\therefore g\left( {2{{x}^{2}}-12x+7} \right)=0\\\\\therefore 2\left( {2{{x}^{2}}-12x+7} \right)+k=0\\\\\therefore 4{{x}^{2}}-24x+14+k=0\\\\\ \text{Since}\left( {g\cdot f} \right)(x)=0\ \text{has two equal roots,}\\\\{{(-24)}^{2}}-4(4)(14+k)=0\\\\\therefore k=22\end{array}$

7.      Let $ \displaystyle f:x\mapsto a + bx, a, b\in R$, be a function from R into R such that $ \displaystyle f(2b)=b$ and $ \displaystyle (f\cdot f)(b) = ab$. If $ \displaystyle f$ is not a constant function, then find the formula for $ \displaystyle f$.

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$ \displaystyle \begin{array}{l}\ \ \ f(x)=a+bx\\\\\ \ \ f(2b)=b\\\\\ \ \ a+b(2b)=b\\\\\therefore a=b-2{{b}^{2}}\\\\\ \ \ \left( {f\cdot f} \right)(b)=ab\\\\\ \ \ f\left( {f(b)} \right)=ab\\\\\ \ \ f\left( {a+{{b}^{2}}} \right)=ab\\\\\ \ \ a+b(a+{{b}^{2}})=ab\\\\\therefore b-2{{b}^{2}}+b(b-2{{b}^{2}}+{{b}^{2}})=(b-2{{b}^{2}})b\\\\\therefore b-2{{b}^{2}}+{{b}^{2}}-{{b}^{3}}={{b}^{2}}-2{{b}^{3}}\\\\\therefore {{b}^{3}}-2{{b}^{2}}+b=0\\\\\therefore b({{b}^{2}}-2b+1)=0\\\\\therefore b{{(b-1)}^{2}}=0\\\\\therefore b=0\ \text{(or)}\ b=1\\\\\text{Since }f(x)\ \text{is not a constant function,}\\b=0\ \text{is impossible}\text{.}\\\\\therefore b=1\\\\\therefore a=(1)-2{{(1)}^{2}}=-1\\\\\therefore f(x)=x-1\end{array}$

8.      $ \displaystyle f:R\to R, g:R\to R$, and $ \displaystyle h:R\to R$ are functions defined by $ \displaystyle f(x) = x^2 + 2, g(x) = x - 1$ and $ \displaystyle h(x) = 3x - 2$. Find the formulae of $ \displaystyle f\cdot g$ and $ \displaystyle f\cdot (h\cdot g)$.

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$ \displaystyle \begin{array}{l}f:R\to R,f(x)={{x}^{2}}+2\\\\g:R\to R,g(x)=x-1\\\\h:R\to R,h(x)=3x-2\\\\\left( {f\cdot g} \right)(x)=f\left( {g(x)} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =f\left( {x-1} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{(x-1)}^{2}}+2\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{x}^{2}}-2x+3\\\\\left( {h\cdot g} \right)(x)=h\left( {g(x)} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =h\left( {x-1} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =3(x-1)-2\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =3x-5\\\\\left( {f\cdot \left( {h\cdot g} \right)} \right)(x)=f\left( {\left( {h\cdot g} \right)(x)} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =f\left( {3x-5} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{(3x-5)}^{2}}+2\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =9{{x}^{2}}-30x+27\end{array}$

9.      The functions $ \displaystyle f$ and $ \displaystyle g$ are defined for all values of $ \displaystyle x$ as follows: $ \displaystyle f:x\mapsto x^2 - 1$ and $ \displaystyle g:x\mapsto (x - 1)^2$.

(i) If $ \displaystyle 4f(x)+3=f(kx)$, find the values of $ \displaystyle k$.

(ii) Express $ \displaystyle g(2x + 1)$ in terms of $ \displaystyle f(x)$.

(iii) Find a function $ \displaystyle h$ such that $ \displaystyle f(x) = g(x) + 2h(x)$.

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$ \displaystyle \begin{array}{*{20}{l}} {f(x)={{x}^{2}}-1,\ g(x)={{{(x-1)}}^{2}}} \\ {} \\ {\text{(i)}\ \ 4f(x)+3=f(kx)} \\ {} \\ {\ \ \ \ \ 4({{x}^{2}}-1)+3={{k}^{2}}{{x}^{2}}-1} \\ {} \\ {\ \ \ \ \ 4{{x}^{2}}-1={{k}^{2}}{{x}^{2}}-1} \\ {} \\ {\ \ \ \ \ \therefore {{k}^{2}}=4\Rightarrow k=\pm 2} \\ {} \\ {\text{(ii)}\ g(x)={{{(x-1)}}^{2}}} \\ {} \\ {\ \ \ \ \ g(2x+1)={{{(2x+1-1)}}^{2}}} \\ {} \\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4{{x}^{2}}} \\ {} \\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4{{x}^{2}}-4+4} \\ {} \\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4({{x}^{2}}-1)+4} \\ {} \\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4\left[ {({{x}^{2}}-1)+1} \right]} \\ {} \\ {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4\left[ {f(x)+1} \right]} \\ {} \\ {\text{(ii)}\ f(x)=g(x)+2h(x)} \\ {} \\ {\ \ \ \ \ {{x}^{2}}-1={{{(x-1)}}^{2}}+2h(x)} \\ {} \\ {\ \ \ \ \ 2h(x)={{x}^{2}}-1-{{{(x-1)}}^{2}}} \\ {} \\ \begin{array}{l}\ \ \ \ \ 2h(x)={{x}^{2}}-1-{{x}^{2}}+2x-1\\\\\ \ \ \ \ \ \ 2h(x)=2x-2\\\\\ \ \ \ \therefore h(x)=x-1\end{array} \end{array}$

Problems Study

 

1. Functions f and g are such that  g^-1 (x) = x  -${\displaystyle \frac{1}{3 }}$ and (f ∘ g ) (x) = 3x - 1.
    Find (g ^-1 ∘ f ) (x) where x ∈ R.

   Solution 

   Let g ^-1(x ) = y then g (y) = x.
   Hence x - ${\displaystyle \frac{1}{3}}$ = y
            x = y + ${\displaystyle \frac{1}{3}}$
            g (y) = y + ${\displaystyle \frac{1}{3}}$
            g (x) = x + ${\displaystyle \frac{1}{3}}$
           (f ∘ g) (x) = 3x - 1
           f (g ) (x)   = 3x - 1
           f (x + ${\displaystyle \frac{1}{3}}$)  = 3x - 1
                          = 3 (x + ${\displaystyle \frac{1}{3}}$) - 2
   Hence f (x ) = 3x - 2
           (g ^-1 ∘ f ) (x) = g ^-1 ( f   (x ) )
                             = g ^-1 ( 3x - 2)
                             = 3x - ${\displaystyle \frac{7}{3}}$

2. If x, y and z are any three consecutive even numbers, Show that x² + y² + z² = 3y² + 8.

   Solution 

   Let x = 2a where a is an integer.

   Since x, y and z are any three consecutive even numbers,

   y = 2a + 2 and z = 2a + 4

   x² + y² + z² = (2a)² + (2a + 2)² + (2a + 4)² 

                     = 4a² + 4a² + 8a + 4 + 4a² + 16a + 16

                     = 12a² + 24a + 20

                     = 12a² + 24a + 12 + 8

                     = 3 (4a² + 8a + 4) + 8 

                     = 3 (2a + 2)² + 8 

                     = 3 y² + 8