\displaystyle \textbf{Question 1: }\text{Find }f+g,\ f-g,\ cf\ (c\in\mathbb{R},\ c\neq0),\ fg,\ \frac{1}{f}\text{ and }\frac{f}{g}
\displaystyle \text{in each of the following:}

\displaystyle \text{(i) }f(x)=x^3+1,\ g(x)=x+1\qquad \text{(ii) }f(x)=\sqrt{x-1},\ g(x)=\sqrt{x+1}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given }f(x)=x^3+1\text{ and }g(x)=x+1
\displaystyle (f+g)(x)=f(x)+g(x)=x^3+1+x+1=x^3+x+2.
\displaystyle \text{Domain}=\mathbb{R}.
\displaystyle (f-g)(x)=f(x)-g(x)=x^3+1-(x+1)=x^3-x.
\displaystyle \text{Domain}=\mathbb{R}.
\displaystyle (cf)(x)=cf(x)=c(x^3+1).
\displaystyle \text{Domain}=\mathbb{R}.
\displaystyle (fg)(x)=f(x)\cdot g(x)=(x^3+1)(x+1)=x^4+x^3+x+1.
\displaystyle \text{Domain}=\mathbb{R}.
\displaystyle \left(\frac{1}{f}\right)(x)=\frac{1}{f(x)}=\frac{1}{x^3+1}.
\displaystyle \text{Domain}=\mathbb{R}-\{-1\}.
\displaystyle \left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{x^3+1}{x+1}=x^2-x+1,\ x\neq-1.
\displaystyle \text{Domain}=\mathbb{R}-\{-1\}.

\displaystyle \text{(ii) Given }f(x)=\sqrt{x-1}\text{ and }g(x)=\sqrt{x+1}
\displaystyle (f+g)(x)=\sqrt{x-1}+\sqrt{x+1}.
\displaystyle \text{Domain}=[1,\infty).
\displaystyle (f-g)(x)=\sqrt{x-1}-\sqrt{x+1}.
\displaystyle \text{Domain}=[1,\infty).
\displaystyle (cf)(x)=c\sqrt{x-1}.
\displaystyle \text{Domain}=[1,\infty).
\displaystyle (fg)(x)=\sqrt{x-1}\cdot\sqrt{x+1}=\sqrt{x^2-1}.
\displaystyle \text{Domain}=[1,\infty).
\displaystyle \left(\frac{1}{f}\right)(x)=\frac{1}{\sqrt{x-1}}.
\displaystyle \text{Domain}=(1,\infty).
\displaystyle \left(\frac{f}{g}\right)(x)=\frac{\sqrt{x-1}}{\sqrt{x+1}}.
\displaystyle \text{Domain}=[1,\infty).
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Let }f(x)=2x+5\text{ and }g(x)=x^2+x.\text{ Describe:}
\displaystyle \text{(i) }f+g\qquad \text{(ii) }f-g\qquad \text{(iii) }fg\qquad \text{(iv) }\frac{f}{g}\text{. Find the domain}
\displaystyle \text{in each case.}
\displaystyle \text{Answer:}
\displaystyle f(x)=2x+5\text{ is defined for every }x\in\mathbb{R}.
\displaystyle \therefore \text{Domain}(f)=\mathbb{R}.
\displaystyle g(x)=x^2+x\text{ is defined for every }x\in\mathbb{R}.
\displaystyle \therefore \text{Domain}(g)=\mathbb{R}.
\displaystyle \therefore \text{Domain}(f)\cap\text{Domain}(g)=\mathbb{R}.

\displaystyle \text{(i) }(f+g)(x)=f(x)+g(x)=2x+5+x^2+x=x^2+3x+5.
\displaystyle \therefore \text{Domain}(f+g)=\mathbb{R}.

\displaystyle \text{(ii) }(f-g)(x)=f(x)-g(x)=2x+5-(x^2+x)=-x^2+x+5.
\displaystyle \therefore \text{Domain}(f-g)=\mathbb{R}.

\displaystyle \text{(iii) }(fg)(x)=f(x)\cdot g(x)=(2x+5)(x^2+x)=2x^3+7x^2+5x.
\displaystyle \therefore \text{Domain}(fg)=\mathbb{R}.

\displaystyle \text{(iv) }\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{2x+5}{x^2+x}.
\displaystyle \text{Since }x^2+x=x(x+1)\neq0,\ x\neq0,-1.
\displaystyle \therefore \text{Domain}\left(\frac{f}{g}\right)=\mathbb{R}-\{0,-1\}.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If }f(x)\text{ is defined on }[-2,2]\text{ by}
\displaystyle f(x)=\left\{\begin{array}{ll}-1,&-2\le x\le0\\x-1,&0<x\le2\end{array}\right. \text{ and }g(x)=f(|x|)+|f(x)|,
\displaystyle \text{find }g(x).
\displaystyle \text{Answer:}
\displaystyle \text{Given }f(x)=\left\{\begin{array}{ll}-1,&-2\le x\le0\\x-1,&0<x\le2\end{array}\right.
\displaystyle \text{Now }g(x)=f(|x|)+|f(x)|.
\displaystyle \text{For }-2\le x\le0,\ |x|=-x,\ f(|x|)=-x-1,\ |f(x)|=1.
\displaystyle \therefore g(x)=(-x-1)+1=-x.
\displaystyle \text{For }0<x\le1,\ f(|x|)=x-1,\ |f(x)|=1-x.
\displaystyle \therefore g(x)=(x-1)+(1-x)=0.
\displaystyle \text{For }1<x\le2,\ f(|x|)=x-1,\ |f(x)|=x-1.
\displaystyle \therefore g(x)=(x-1)+(x-1)=2x-2.
\displaystyle \therefore g(x)=\left\{\begin{array}{ll}-x,&-2\le x\le0\\0,&0<x\le1\\2x-2,&1<x\le2\end{array}\right.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Let }f\text{ and }g\text{ be two real functions defined by }f(x)=\sqrt{x+1}
\displaystyle \text{and }g(x)=\sqrt{9-x^2}.\text{ Describe each of the following functions:}
\displaystyle \text{(i) }f+g\qquad \text{(ii) }g-f\qquad \text{(iii) }fg\qquad \text{(iv) }\frac{f}{g}\qquad \text{(v) }\frac{g}{f}
\displaystyle \text{(vi) }2f-\sqrt{5}g\qquad \text{(vii) }f^2+7f\qquad \text{(viii) }\frac{5}{g}
\displaystyle \text{Answer:}
\displaystyle \text{Given }f(x)=\sqrt{x+1}\text{ and }g(x)=\sqrt{9-x^2}.
\displaystyle f(x)\text{ is defined when }x+1\geq0.
\displaystyle \Rightarrow x\geq-1
\displaystyle \therefore \text{Domain}(f)=[-1,\infty).
\displaystyle g(x)\text{ is defined when }9-x^2\geq0.
\displaystyle \Rightarrow (3-x)(3+x)\geq0
\displaystyle \Rightarrow -3\leq x\leq3
\displaystyle \therefore \text{Domain}(g)=[-3,3].
\displaystyle \text{Therefore, }\text{Domain}(f)\cap\text{Domain}(g)=[-1,3].

\displaystyle \text{(i) }(f+g)(x)=f(x)+g(x)=\sqrt{x+1}+\sqrt{9-x^2}.
\displaystyle \therefore f+g:[-1,3]\rightarrow\mathbb{R}.

\displaystyle \text{(ii) }(g-f)(x)=g(x)-f(x)=\sqrt{9-x^2}-\sqrt{x+1}.
\displaystyle \therefore g-f:[-1,3]\rightarrow\mathbb{R}.

\displaystyle \text{(iii) }(fg)(x)=f(x)g(x)=\sqrt{x+1}\sqrt{9-x^2}.
\displaystyle \therefore fg:[-1,3]\rightarrow\mathbb{R}.

\displaystyle \text{(iv) }\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{\sqrt{x+1}}{\sqrt{9-x^2}}.
\displaystyle \text{Here }g(x)\neq0,\text{ so }9-x^2>0.
\displaystyle \text{Combining this with }x\geq-1,\text{ we get }-1\leq x<3.
\displaystyle \therefore \frac{f}{g}:[-1,3)\rightarrow\mathbb{R}.

\displaystyle \text{(v) }\left(\frac{g}{f}\right)(x)=\frac{g(x)}{f(x)}=\frac{\sqrt{9-x^2}}{\sqrt{x+1}}.
\displaystyle \text{Here }f(x)\neq0,\text{ so }x+1>0.
\displaystyle \text{Combining this with }x\leq3,\text{ we get }-1<x\leq3.
\displaystyle \therefore \frac{g}{f}:(-1,3]\rightarrow\mathbb{R}.

\displaystyle \text{(vi) }(2f-\sqrt{5}g)(x)=2f(x)-\sqrt{5}g(x)
\displaystyle =2\sqrt{x+1}-\sqrt{5}\sqrt{9-x^2}
\displaystyle =2\sqrt{x+1}-\sqrt{45-5x^2}.
\displaystyle \therefore 2f-\sqrt{5}g:[-1,3]\rightarrow\mathbb{R}.

\displaystyle \text{(vii) }(f^2+7f)(x)=[f(x)]^2+7f(x)
\displaystyle =(\sqrt{x+1})^2+7\sqrt{x+1}
\displaystyle =x+1+7\sqrt{x+1}.
\displaystyle \therefore f^2+7f:[-1,\infty)\rightarrow\mathbb{R}.

\displaystyle \text{(viii) }\left(\frac{5}{g}\right)(x)=\frac{5}{g(x)}=\frac{5}{\sqrt{9-x^2}}.
\displaystyle \text{Since the square root is in the denominator, }9-x^2>0.
\displaystyle \Rightarrow -3<x<3
\displaystyle \therefore \frac{5}{g}:(-3,3)\rightarrow\mathbb{R}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }f(x)=\log_e(1-x)\text{ and }g(x)=[x],\text{ determine each of the}
\displaystyle \text{following functions:}
\displaystyle \text{(i) }f+g\qquad \text{(ii) }fg\qquad \text{(iii) }g-f\qquad \text{(iv) }\frac{f}{g}\qquad \text{(v) }\frac{g}{f}
\displaystyle \text{(vi) }(f+g)(-1),\ (fg)(0),\ \left(\frac{f}{g}\right)\left(\frac{1}{2}\right),\ \left(\frac{g}{f}\right)\left(\frac{1}{2}\right).
\displaystyle \text{Answer:}
\displaystyle \text{Given }f(x)=\log_e(1-x)\text{ and }g(x)=[x].
\displaystyle f(x)\text{ is defined when }1-x>0.
\displaystyle \Rightarrow x<1
\displaystyle \therefore \text{Domain}(f)=(-\infty,1).
\displaystyle g(x)=[x]\text{ is defined for every }x\in\mathbb{R}.
\displaystyle \therefore \text{Domain}(g)=\mathbb{R}.
\displaystyle \therefore \text{Domain}(f)\cap\text{Domain}(g)=(-\infty,1).

\displaystyle \text{(i) }(f+g)(x)=f(x)+g(x)=\log_e(1-x)+[x].
\displaystyle \therefore f+g:(-\infty,1)\rightarrow\mathbb{R}.

\displaystyle \text{(ii) }(fg)(x)=f(x)g(x)=[x]\log_e(1-x).
\displaystyle \therefore fg:(-\infty,1)\rightarrow\mathbb{R}.

\displaystyle \text{(iii) }(g-f)(x)=g(x)-f(x)=[x]-\log_e(1-x).
\displaystyle \therefore g-f:(-\infty,1)\rightarrow\mathbb{R}.

\displaystyle \text{(iv) }\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{\log_e(1-x)}{[x]}.
\displaystyle \text{For the quotient to be defined, }x<1\text{ and }[x]\neq0.
\displaystyle [x]=0\text{ for }0\leq x<1.
\displaystyle \therefore \text{Domain}\left(\frac{f}{g}\right)=(-\infty,0).
\displaystyle \therefore \frac{f}{g}:(-\infty,0)\rightarrow\mathbb{R}.

\displaystyle \text{(v) }\left(\frac{g}{f}\right)(x)=\frac{g(x)}{f(x)}=\frac{[x]}{\log_e(1-x)}.
\displaystyle \text{For the quotient to be defined, }x<1\text{ and }\log_e(1-x)\neq0.
\displaystyle \log_e(1-x)=0
\displaystyle \Rightarrow 1-x=1
\displaystyle \Rightarrow x=0
\displaystyle \therefore \text{Domain}\left(\frac{g}{f}\right)=(-\infty,0)\cup(0,1).
\displaystyle \therefore \frac{g}{f}:(-\infty,0)\cup(0,1)\rightarrow\mathbb{R}.

\displaystyle \text{(vi) }(f+g)(-1)=f(-1)+g(-1)
\displaystyle =\log_e(1-(-1))+[-1]
\displaystyle =\log_e2-1.
\displaystyle (fg)(0)=f(0)g(0)
\displaystyle =\log_e(1-0)[0]=0.
\displaystyle \left(\frac{f}{g}\right)\left(\frac{1}{2}\right)=\frac{\log_e\left(1-\frac{1}{2}\right)}{\left[\frac{1}{2}\right]}.
\displaystyle \text{Since }\left[\frac{1}{2}\right]=0,\ \left(\frac{f}{g}\right)\left(\frac{1}{2}\right)\text{ is not defined.}
\displaystyle \left(\frac{g}{f}\right)\left(\frac{1}{2}\right)=\frac{\left[\frac{1}{2}\right]}{\log_e\left(1-\frac{1}{2}\right)}
\displaystyle =\frac{0}{\log_e\left(\frac{1}{2}\right)}=0.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If }f,\ g,\ h\text{ are real functions defined by }f(x)=\sqrt{x+1},
\displaystyle g(x)=\frac{1}{x}\text{ and }h(x)=2x^2-3,\text{ then find the values of }
\displaystyle (2f+g-h)(1)\text{ and }(2f+g-h)(0).
\displaystyle \text{Answer:}
\displaystyle \text{Given }f(x)=\sqrt{x+1},\ g(x)=\frac{1}{x}\text{ and }h(x)=2x^2-3.
\displaystyle f(x)\text{ is defined for }x\ge-1.
\displaystyle \therefore \text{Domain}(f)=[-1,\infty).
\displaystyle g(x)\text{ is defined for }x\neq0.
\displaystyle \therefore \text{Domain}(g)=\mathbb{R}-\{0\}.
\displaystyle h(x)\text{ is defined for every }x\in\mathbb{R}.
\displaystyle \therefore \text{Domain}(h)=\mathbb{R}.
\displaystyle \therefore \text{Domain}(f)\cap\text{Domain}(g)\cap\text{Domain}(h)=[-1,\infty)-\{0\}.
\displaystyle \therefore (2f+g-h)(x)=2f(x)+g(x)-h(x).
\displaystyle =2\sqrt{x+1}+\frac{1}{x}-(2x^2-3).
\displaystyle =2\sqrt{x+1}+\frac{1}{x}-2x^2+3.
\displaystyle (2f+g-h)(1)=2\sqrt{2}+\frac{1}{1}-2(1)^2+3
\displaystyle =2\sqrt{2}+1-2+3=2\sqrt{2}+2.
\displaystyle (2f+g-h)(0)\text{ is not defined since }0\notin\text{Domain}(2f+g-h).
\displaystyle \\

Question 7: The function f is defined by 

\displaystyle f(x) = \Bigg\{ \begin{array}{lll} 1-x, \hspace*{1.0cm} x < 0 \\ 1, \hspace*{1.7cm} x = 0 \\ x+1, \hspace*{1.0cm} x > 0 \end{array}

Draw the graph of \displaystyle f(x)

Answer:

\displaystyle \text{Given }  f(x) = \Bigg\{ \begin{array}{lll} 1-x, \hspace*{1.0cm} x < 0 \\ 1, \hspace*{1.7cm} x = 0 \\ x+1, \hspace*{1.0cm} x > 0 \end{array}

\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Let }f,g:\mathbb{R}\rightarrow\mathbb{R}\text{ be defined respectively by }
\displaystyle f(x)=x+1,\ g(x)=2x-3.\text{ Find }f+g,\ f-g\text{ and }\frac{f}{g}.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f(x)=x+1,\ g(x)=2x-3.
\displaystyle f(x)\text{ is defined for every }x\in\mathbb{R}.
\displaystyle \therefore \text{Domain}(f)=\mathbb{R}.
\displaystyle g(x)\text{ is defined for every }x\in\mathbb{R}.
\displaystyle \therefore \text{Domain}(g)=\mathbb{R}.
\displaystyle \therefore \text{Domain}(f)\cap\text{Domain}(g)=\mathbb{R}.
\displaystyle (f+g)(x)=f(x)+g(x)=(x+1)+(2x-3)=3x-2.
\displaystyle \therefore f+g:\mathbb{R}\rightarrow\mathbb{R}.
\displaystyle (f-g)(x)=f(x)-g(x)=(x+1)-(2x-3)=-x+4.
\displaystyle \therefore f-g:\mathbb{R}\rightarrow\mathbb{R}.
\displaystyle \frac{1}{g(x)}=\frac{1}{2x-3}\text{ is defined when }2x-3\neq0.
\displaystyle \Rightarrow x\neq\frac{3}{2}.
\displaystyle \therefore \text{Domain}\left(\frac{f}{g}\right)=\mathbb{R}-\left\{\frac{3}{2}\right\}.
\displaystyle \left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{x+1}{2x-3}.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Let }f:[0,\infty)\rightarrow\mathbb{R}\text{ and }g:\mathbb{R}\rightarrow\mathbb{R}
\displaystyle \text{be defined by }f(x)=\sqrt{x}\text{ and }g(x)=x.\text{ Find }f+g,\ f-g,\ fg\text{ and }\frac{f}{g}.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:[0,\infty)\rightarrow\mathbb{R},\ f(x)=\sqrt{x}.
\displaystyle g:\mathbb{R}\rightarrow\mathbb{R},\ g(x)=x.
\displaystyle \therefore \text{Domain}(f)\cap\text{Domain}(g)=[0,\infty).
\displaystyle (f+g)(x)=f(x)+g(x)=\sqrt{x}+x.
\displaystyle \therefore f+g:[0,\infty)\rightarrow\mathbb{R}.
\displaystyle (f-g)(x)=f(x)-g(x)=\sqrt{x}-x.
\displaystyle \therefore f-g:[0,\infty)\rightarrow\mathbb{R}.
\displaystyle (fg)(x)=f(x)\cdot g(x)=x\sqrt{x}.
\displaystyle \therefore fg:[0,\infty)\rightarrow\mathbb{R}.
\displaystyle \left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{\sqrt{x}}{x}=\frac{1}{\sqrt{x}},\ x>0.
\displaystyle \therefore \text{Domain}\left(\frac{f}{g}\right)=(0,\infty).
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Let }f(x)=x^2\text{ and }g(x)=2x+1\text{ be two real functions. Find }
\displaystyle (f+g)(x),\ (f-g)(x),\ (fg)(x)\text{ and }\left(\frac{f}{g}\right)(x).
\displaystyle \text{Answer:}
\displaystyle \text{Given }f(x)=x^2\text{ and }g(x)=2x+1.
\displaystyle f(x)\text{ is defined for every }x\in\mathbb{R}.
\displaystyle \therefore \text{Domain}(f)=\mathbb{R}.
\displaystyle g(x)\text{ is defined for every }x\in\mathbb{R}.
\displaystyle \therefore \text{Domain}(g)=\mathbb{R}.
\displaystyle \therefore \text{Domain}(f)\cap\text{Domain}(g)=\mathbb{R}.
\displaystyle (f+g)(x)=f(x)+g(x)=x^2+2x+1.
\displaystyle \therefore f+g:\mathbb{R}\rightarrow\mathbb{R}.
\displaystyle (f-g)(x)=f(x)-g(x)=x^2-2x-1.
\displaystyle \therefore f-g:\mathbb{R}\rightarrow\mathbb{R}.
\displaystyle (fg)(x)=f(x)\cdot g(x)=x^2(2x+1).
\displaystyle \therefore fg:\mathbb{R}\rightarrow\mathbb{R}.
\displaystyle \left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{x^2}{2x+1}.
\displaystyle \text{Since }2x+1\neq0,\ x\neq-\frac12.
\displaystyle \therefore \text{Domain}\left(\frac{f}{g}\right)=\mathbb{R}-\left\{-\frac12\right\}.
\displaystyle \\

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