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$.

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$ \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$.

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

$ \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)$.

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$ \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 $\displaystyle q$.

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$ \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}$