\displaystyle \textbf{Question 1: }\text{Find }f\circ g\text{ and }g\circ f,\text{ if}
\displaystyle \text{(i) }f(x)=e^x,\ g(x)=\log_e x.
\displaystyle \text{Answer:}
\displaystyle \text{Let }f:\mathbb R\rightarrow(0,\infty)\text{ and }g:(0,\infty)\rightarrow\mathbb R.
\displaystyle \text{Since Range}(g)=\mathbb R=\text{Domain}(f),\ f\circ g\text{ exists.}
\displaystyle (f\circ g)(x)=f(g(x))=f(\log_e x)=e^{\log_e x}=x,\qquad x>0.
\displaystyle \therefore f\circ g:(0,\infty)\rightarrow(0,\infty)\text{ is given by }(f\circ g)(x)=x.
\displaystyle \text{Since Range}(f)=(0,\infty)=\text{Domain}(g),\ g\circ f\text{ exists.}
\displaystyle (g\circ f)(x)=g(f(x))=g(e^x)=\log_e(e^x)=x,\qquad x\in\mathbb R.
\displaystyle \therefore g\circ f:\mathbb R\rightarrow\mathbb R\text{ is given by }(g\circ f)(x)=x.
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Find }f\circ g\text{ and }g\circ f,\text{ if}
\displaystyle \text{(ii) }f(x)=x^2,\ g(x)=\cos x.
\displaystyle \text{Answer:}
\displaystyle \text{Let }f:\mathbb R\rightarrow[0,\infty)\text{ and }g:\mathbb R\rightarrow[-1,1].
\displaystyle \text{Since Range}(g)=[-1,1]\subseteq\text{Domain}(f)=\mathbb R,\ f\circ g\text{ exists.}
\displaystyle (f\circ g)(x)=f(g(x))=f(\cos x)=(\cos x)^2=\cos^2x.
\displaystyle \therefore f\circ g:\mathbb R\rightarrow\mathbb R\text{ is given by }(f\circ g)(x)=\cos^2x.
\displaystyle \text{Since Range}(f)=[0,\infty)\subseteq\text{Domain}(g)=\mathbb R,\ g\circ f\text{ exists.}
\displaystyle (g\circ f)(x)=g(f(x))=g(x^2)=\cos(x^2).
\displaystyle \therefore g\circ f:\mathbb R\rightarrow\mathbb R\text{ is given by }(g\circ f)(x)=\cos(x^2).
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Find }f\circ g\text{ and }g\circ f,\text{ if}
\displaystyle \text{(iii) }f(x)=|x|,\ g(x)=\sin x.
\displaystyle \text{Answer:}
\displaystyle \text{Let }f:\mathbb R\rightarrow[0,\infty)\text{ and }g:\mathbb R\rightarrow[-1,1].
\displaystyle \text{Since Range}(g)=[-1,1]\subseteq\text{Domain}(f)=\mathbb R,\ f\circ g\text{ exists.}
\displaystyle (f\circ g)(x)=f(g(x))=f(\sin x)=|\sin x|.
\displaystyle \therefore f\circ g:\mathbb R\rightarrow\mathbb R\text{ is given by }(f\circ g)(x)=|\sin x|.
\displaystyle \text{Since Range}(f)=[0,\infty)\subseteq\text{Domain}(g)=\mathbb R,\ g\circ f\text{ exists.}
\displaystyle (g\circ f)(x)=g(f(x))=g(|x|)=\sin|x|.
\displaystyle \therefore g\circ f:\mathbb R\rightarrow\mathbb R\text{ is given by }(g\circ f)(x)=\sin|x|.
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Find }f\circ g\text{ and }g\circ f,\text{ if}
\displaystyle \text{(iv) }f(x)=x+1,\ g(x)=e^x.
\displaystyle \text{Answer:}
\displaystyle \text{Let }f:\mathbb R\rightarrow\mathbb R\text{ and }g:\mathbb R\rightarrow(0,\infty).
\displaystyle \text{Since Range}(g)=(0,\infty)\subseteq\text{Domain}(f)=\mathbb R,\ f\circ g\text{ exists.}
\displaystyle (f\circ g)(x)=f(g(x))=f(e^x)=e^x+1.
\displaystyle \therefore f\circ g:\mathbb R\rightarrow\mathbb R\text{ is given by }(f\circ g)(x)=e^x+1.
\displaystyle \text{Since Range}(f)=\mathbb R=\text{Domain}(g),\ g\circ f\text{ exists.}
\displaystyle (g\circ f)(x)=g(f(x))=g(x+1)=e^{x+1}.
\displaystyle \therefore g\circ f:\mathbb R\rightarrow\mathbb R\text{ is given by }(g\circ f)(x)=e^{x+1}.
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Find }f\circ g\text{ and }g\circ f,\text{ if}
\displaystyle \text{(v) }f(x)=\sin^{-1}x,\ g(x)=x^2.
\displaystyle \text{Answer:}
\displaystyle \text{Let }f:[-1,1]\rightarrow\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\text{ and }g:\mathbb R\rightarrow[0,\infty).
\displaystyle \text{Since Range}(g)\nsubseteq\text{Domain}(f),\text{ the domain of }f\circ g\text{ is}
\displaystyle \text{Domain}(f\circ g)=\{x:x\in\mathbb R\text{ and }x^2\in[-1,1]\}=[-1,1].
\displaystyle \therefore f\circ g:[-1,1]\rightarrow\mathbb R\text{ is given by}
\displaystyle (f\circ g)(x)=f(g(x))=f(x^2)=\sin^{-1}(x^2).
\displaystyle \text{Since Range}(f)=\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\subseteq\text{Domain}(g)=\mathbb R,
\displaystyle g\circ f\text{ exists.}
\displaystyle (g\circ f)(x)=g(f(x))=g(\sin^{-1}x)=(\sin^{-1}x)^2.
\displaystyle \therefore g\circ f:[-1,1]\rightarrow\mathbb R\text{ is given by }(g\circ f)(x)=(\sin^{-1}x)^2.
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Find }f\circ g\text{ and }g\circ f,\text{ if}
\displaystyle \text{(vi) }f(x)=x+1,\ g(x)=\sin x.
\displaystyle \text{Answer:}
\displaystyle \text{Let }f:\mathbb R\rightarrow\mathbb R\text{ and }g:\mathbb R\rightarrow[-1,1].
\displaystyle \text{Since Range}(g)=[-1,1]\subseteq\text{Domain}(f)=\mathbb R,\ f\circ g\text{ exists.}
\displaystyle (f\circ g)(x)=f(g(x))=f(\sin x)=\sin x+1.
\displaystyle \therefore f\circ g:\mathbb R\rightarrow\mathbb R\text{ is given by }(f\circ g)(x)=\sin x+1.
\displaystyle \text{Since Range}(f)=\mathbb R=\text{Domain}(g),\ g\circ f\text{ exists.}
\displaystyle (g\circ f)(x)=g(f(x))=g(x+1)=\sin(x+1).
\displaystyle \therefore g\circ f:\mathbb R\rightarrow\mathbb R\text{ is given by }(g\circ f)(x)=\sin(x+1).
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Find }f\circ g\text{ and }g\circ f,\text{ if}
\displaystyle \text{(vii) }f(x)=x+1,\ g(x)=2x+3.
\displaystyle \text{Answer:}
\displaystyle \text{Let }f:\mathbb R\rightarrow\mathbb R\text{ and }g:\mathbb R\rightarrow\mathbb R.
\displaystyle \text{Since Range}(g)\subseteq\text{Domain}(f),\ f\circ g\text{ exists.}
\displaystyle (f\circ g)(x)=f(g(x))=f(2x+3)=2x+3+1=2x+4.
\displaystyle \therefore f\circ g:\mathbb R\rightarrow\mathbb R\text{ is given by }(f\circ g)(x)=2x+4.
\displaystyle \text{Since Range}(f)\subseteq\text{Domain}(g),\ g\circ f\text{ exists.}
\displaystyle (g\circ f)(x)=g(f(x))=g(x+1)=2(x+1)+3=2x+5.
\displaystyle \therefore g\circ f:\mathbb R\rightarrow\mathbb R\text{ is given by }(g\circ f)(x)=2x+5.
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Find }f\circ g\text{ and }g\circ f,\text{ if}
\displaystyle \text{(viii) }f(x)=c,\ c\in\mathbb R,\ g(x)=\sin(x^2).
\displaystyle \text{Answer:}
\displaystyle \text{Let }f:\mathbb R\rightarrow\{c\}\text{ and }g:\mathbb R\rightarrow[-1,1].
\displaystyle \text{Since Range}(g)=[-1,1]\subseteq\text{Domain}(f)=\mathbb R,\ f\circ g\text{ exists.}
\displaystyle (f\circ g)(x)=f(g(x))=f(\sin(x^2))=c.
\displaystyle \therefore f\circ g:\mathbb R\rightarrow\{c\}\text{ is given by }(f\circ g)(x)=c.
\displaystyle \text{Since Range}(f)=\{c\}\subseteq\text{Domain}(g)=\mathbb R,\ g\circ f\text{ exists.}
\displaystyle (g\circ f)(x)=g(f(x))=g(c)=\sin(c^2).
\displaystyle \therefore g\circ f:\mathbb R\rightarrow\mathbb R\text{ is given by }(g\circ f)(x)=\sin(c^2).
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Find }f\circ g\text{ and }g\circ f,\text{ if}
\displaystyle \text{(ix) }f(x)=x^2+2,\quad g(x)=1-\frac{1}{1-x}.
\displaystyle \text{Answer:}
\displaystyle f:\mathbb R\rightarrow[2,\infty).
\displaystyle \text{For the domain of }g,\quad 1-x\ne0\Rightarrow x\ne1.
\displaystyle \therefore \text{Domain}(g)=\mathbb R-\{1\}.
\displaystyle g(x)=1-\frac{1}{1-x}=\frac{1-x-1}{1-x}=\frac{-x}{1-x}=\frac{x}{x-1}.
\displaystyle \text{To find the range of }g,\text{ let }y=\frac{x}{x-1}.
\displaystyle yx-y=x\Rightarrow x(y-1)=y\Rightarrow x=\frac{y}{y-1}.
\displaystyle \text{Thus, }y\ne1.
\displaystyle \therefore \text{Range}(g)=\mathbb R-\{1\}.
\displaystyle \therefore g:\mathbb R-\{1\}\rightarrow\mathbb R-\{1\}.
\displaystyle \text{Since Range}(g)\subseteq\text{Domain}(f),\ f\circ g\text{ exists.}
\displaystyle \text{Domain}(f\circ g)=\mathbb R-\{1\}.
\displaystyle (f\circ g)(x)=f(g(x))=f\left(\frac{x}{x-1}\right).
\displaystyle =\left(\frac{x}{x-1}\right)^2+2.
\displaystyle =\frac{x^2+2(x-1)^2}{(x-1)^2}.
\displaystyle =\frac{3x^2-4x+2}{(x-1)^2}.
\displaystyle \therefore f\circ g:\mathbb R-\{1\}\rightarrow[2,\infty)\text{ is given by}
\displaystyle (f\circ g)(x)=\frac{3x^2-4x+2}{(x-1)^2}.
\displaystyle \text{Since Range}(f)=[2,\infty)\subseteq\text{Domain}(g)=\mathbb R-\{1\},
\displaystyle g\circ f\text{ exists and its domain is }\mathbb R.
\displaystyle (g\circ f)(x)=g(f(x))=g(x^2+2).
\displaystyle =1-\frac{1}{1-(x^2+2)}.
\displaystyle =1+\frac{1}{x^2+1}=\frac{x^2+2}{x^2+1}.
\displaystyle \therefore g\circ f:\mathbb R\rightarrow(1,2]\text{ is given by}
\displaystyle (g\circ f)(x)=\frac{x^2+2}{x^2+1}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Let }f(x)=x^2+x+1\text{ and }g(x)=\sin x.\text{ Show that }f\circ g\ne g\circ f.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f(x)=x^2+x+1\text{ and }g(x)=\sin x.
\displaystyle (f\circ g)(x)=f(g(x))=f(\sin x)=\sin^2x+\sin x+1.
\displaystyle (g\circ f)(x)=g(f(x))=g(x^2+x+1)=\sin(x^2+x+1).
\displaystyle \text{At }x=0,
\displaystyle (f\circ g)(0)=1\text{ and }(g\circ f)(0)=\sin1.
\displaystyle \text{Since }1\ne\sin1,
\displaystyle \therefore f\circ g\ne g\circ f.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If }f(x)=|x|,\text{ prove that }f\circ f=f.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f(x)=|x|.
\displaystyle (f\circ f)(x)=f(f(x))=f(|x|)=||x||=|x|=f(x).
\displaystyle \therefore (f\circ f)(x)=f(x),\ \forall x\in\mathbb R.
\displaystyle \therefore f\circ f=f.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }f(x)=2x+5\text{ and }g(x)=x^2+1\text{ are two real functions, describe}
\displaystyle \text{(i) }f\circ g\qquad\text{(ii) }g\circ f\qquad\text{(iii) }f\circ f\qquad\text{(iv) }f^2.
\displaystyle \text{Answer:}
\displaystyle \text{Since }f\text{ and }g\text{ are polynomial functions, }f:\mathbb R\rightarrow\mathbb R\text{ and }g:\mathbb R\rightarrow\mathbb R.
\displaystyle \text{Hence, all the required compositions exist.}
\displaystyle \text{(i) }(f\circ g)(x)=f(g(x))=f(x^2+1)=2(x^2+1)+5=2x^2+7.
\displaystyle \text{(ii) }(g\circ f)(x)=g(f(x))=g(2x+5)=(2x+5)^2+1=4x^2+20x+26.
\displaystyle \text{(iii) }(f\circ f)(x)=f(f(x))=f(2x+5)=2(2x+5)+5=4x+15.
\displaystyle \text{(iv) }f^2(x)=f(x)\times f(x)=(2x+5)^2=4x^2+20x+25.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }f(x)=\sin x\text{ and }g(x)=2x\text{ are two real functions, describe }g\circ f
\displaystyle \text{ and }f\circ g.\text{ Are these equal functions?}
\displaystyle \text{Answer:}
\displaystyle \text{Given }f(x)=\sin x\text{ and }g(x)=2x.
\displaystyle \text{Here }f:\mathbb R\rightarrow[-1,1]\text{ and }g:\mathbb R\rightarrow\mathbb R.
\displaystyle \text{Since Range}(f)=[-1,1]\subseteq\text{Domain}(g)=\mathbb R,\ g\circ f\text{ exists.}
\displaystyle (g\circ f)(x)=g(f(x))=g(\sin x)=2\sin x.
\displaystyle \text{Since Range}(g)=\mathbb R=\text{Domain}(f),\ f\circ g\text{ exists.}
\displaystyle (f\circ g)(x)=f(g(x))=f(2x)=\sin(2x).
\displaystyle \text{At }x=\frac{\pi}{4},\ (g\circ f)\!\left(\frac{\pi}{4}\right)=\sqrt2,\ (f\circ g)\!\left(\frac{\pi}{4}\right)=1.
\displaystyle \text{Since }\sqrt2\ne1,
\displaystyle \therefore f\circ g\ne g\circ f.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Let }f,g,h\text{ be real functions given by }f(x)=\sin x,\ g(x)=2x
\displaystyle \text{and }h(x)=\cos x.\text{ Prove that }f\circ g=g\circ(fh).
\displaystyle \text{Answer:}
\displaystyle \text{Given }f(x)=\sin x,\quad g(x)=2x,\quad h(x)=\cos x.
\displaystyle \text{Here }f:\mathbb R\rightarrow[-1,1]\text{ and }g:\mathbb R\rightarrow\mathbb R.
\displaystyle \text{Since Range}(g)=\mathbb R=\text{Domain}(f),\ f\circ g\text{ exists.}
\displaystyle \text{Also, }(fh)(x)=f(x)h(x)=\sin x\cos x=\frac12\sin(2x).
\displaystyle \text{Since }-1\le\sin(2x)\le1,
\displaystyle -\frac12\le\frac12\sin(2x)\le\frac12.
\displaystyle \therefore \text{Range}(fh)=\left[-\frac12,\frac12\right].
\displaystyle \text{Since Range}(fh)\subseteq\text{Domain}(g)=\mathbb R,\ g\circ(fh)\text{ exists.}
\displaystyle \text{Thus, both }f\circ g\text{ and }g\circ(fh)\text{ have domain }\mathbb R.
\displaystyle (f\circ g)(x)=f(g(x))=f(2x)=\sin(2x).
\displaystyle (g\circ(fh))(x)=g((fh)(x))=g(\sin x\cos x).
\displaystyle =2\sin x\cos x=\sin(2x).
\displaystyle \therefore (f\circ g)(x)=(g\circ(fh))(x),\ \forall x\in\mathbb R.
\displaystyle \therefore f\circ g=g\circ(fh).
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Let }f\text{ be any real function and let }g\text{ be defined by }g(x)=2x.
\displaystyle \text{Prove that }g\circ f=f+f.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f:\mathbb R\rightarrow\mathbb R.
\displaystyle \text{Since }g(x)=2x,\ g:\mathbb R\rightarrow\mathbb R.
\displaystyle \text{Hence, }g\circ f\text{ and }f+f\text{ have the same domain }\mathbb R.
\displaystyle (g\circ f)(x)=g(f(x))=2f(x).
\displaystyle (f+f)(x)=f(x)+f(x)=2f(x).
\displaystyle \therefore (g\circ f)(x)=(f+f)(x),\ \forall x\in\mathbb R.
\displaystyle \therefore g\circ f=f+f.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If }f(x)=\sqrt{1-x}\text{ and }g(x)=\log_e x\text{ are two real functions,}
\displaystyle \text{describe the functions }f\circ g\text{ and }g\circ f.
\displaystyle \text{Answer:}
\displaystyle f(x)=\sqrt{1-x}.
\displaystyle \text{For the domain of }f,\quad 1-x\ge0.
\displaystyle \Rightarrow x\le1.
\displaystyle \therefore \text{Domain}(f)=(-\infty,1].
\displaystyle \text{Also, }\sqrt{1-x}\ge0\text{ and it can take every non-negative real value.}
\displaystyle \therefore \text{Range}(f)=[0,\infty).
\displaystyle \therefore f:(-\infty,1]\rightarrow[0,\infty).
\displaystyle g(x)=\log_e x.
\displaystyle \therefore g:(0,\infty)\rightarrow\mathbb R.
\displaystyle \text{Now, Range}(g)=\mathbb R\nsubseteq\text{Domain}(f)=(-\infty,1].
\displaystyle \therefore \text{Domain}(f\circ g)=\{x:x\in(0,\infty)\text{ and }\log_e x\le1\}.
\displaystyle \log_e x\le1\Rightarrow x\le e.
\displaystyle \therefore \text{Domain}(f\circ g)=(0,e].
\displaystyle (f\circ g)(x)=f(g(x))=f(\log_e x)=\sqrt{1-\log_e x}.
\displaystyle \therefore f\circ g:(0,e]\rightarrow[0,\infty)\text{ is given by}
\displaystyle (f\circ g)(x)=\sqrt{1-\log_e x}.
\displaystyle \text{Now, Range}(f)=[0,\infty)\nsubseteq\text{Domain}(g)=(0,\infty).
\displaystyle \therefore \text{Domain}(g\circ f)=\{x:x\le1\text{ and }\sqrt{1-x}>0\}.
\displaystyle \sqrt{1-x}>0\Rightarrow 1-x>0\Rightarrow x<1.
\displaystyle \therefore \text{Domain}(g\circ f)=(-\infty,1).
\displaystyle (g\circ f)(x)=g(f(x))=g(\sqrt{1-x}).
\displaystyle =\log_e\sqrt{1-x}=\log_e(1-x)^{\frac12}=\frac12\log_e(1-x).
\displaystyle \therefore g\circ f:(-\infty,1)\rightarrow\mathbb R\text{ is given by}
\displaystyle (g\circ f)(x)=\frac12\log_e(1-x).
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If }f:\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\rightarrow\mathbb R\text{ and }g:[-1,1]\rightarrow\mathbb R
\displaystyle \text{are defined by }f(x)=\tan x\text{ and }g(x)=\sqrt{1-x^2},\text{ respectively, describe }f\circ g\text{ and }g\circ f.
\displaystyle \text{Answer:}
\displaystyle g(x)=\sqrt{1-x^2},\qquad x\in[-1,1].
\displaystyle \text{Since }0\le x^2\le1,
\displaystyle 0\le1-x^2\le1.
\displaystyle \Rightarrow 0\le\sqrt{1-x^2}\le1.
\displaystyle \therefore \text{Range}(g)=[0,1].
\displaystyle \text{Thus, }f:\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\rightarrow\mathbb R\text{ and }g:[-1,1]\rightarrow[0,1].
\displaystyle \text{Since Range}(g)=[0,1]\subseteq\text{Domain}(f)=\left(-\frac{\pi}{2},\frac{\pi}{2}\right),
\displaystyle f\circ g\text{ exists and its domain is }[-1,1].
\displaystyle (f\circ g)(x)=f(g(x))=f\left(\sqrt{1-x^2}\right).
\displaystyle =\tan\left(\sqrt{1-x^2}\right).
\displaystyle \therefore f\circ g:[-1,1]\rightarrow\mathbb R\text{ is given by}
\displaystyle (f\circ g)(x)=\tan\left(\sqrt{1-x^2}\right).
\displaystyle \text{Now, Range}(f)=\mathbb R\nsubseteq\text{Domain}(g)=[-1,1].
\displaystyle \therefore \text{Domain}(g\circ f)=\left\{x:x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\text{ and }\tan x\in[-1,1]\right\}.
\displaystyle \text{Since }\tan x\text{ is increasing on }\left(-\frac{\pi}{2},\frac{\pi}{2}\right),
\displaystyle -1\le\tan x\le1\Rightarrow-\frac{\pi}{4}\le x\le\frac{\pi}{4}.
\displaystyle \therefore \text{Domain}(g\circ f)=\left[-\frac{\pi}{4},\frac{\pi}{4}\right].
\displaystyle (g\circ f)(x)=g(f(x))=g(\tan x)=\sqrt{1-\tan^2x}.
\displaystyle \therefore g\circ f:\left[-\frac{\pi}{4},\frac{\pi}{4}\right]\rightarrow[0,1]\text{ is given by}
\displaystyle (g\circ f)(x)=\sqrt{1-\tan^2x}.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }f(x)=\sqrt{x+3}\text{ and }g(x)=x^2+1\text{ are two real functions,}
\displaystyle \text{find }f\circ g\text{ and }g\circ f.
\displaystyle \text{Answer:}
\displaystyle f(x)=\sqrt{x+3}.
\displaystyle \text{For the domain of }f,\ x+3\ge0.
\displaystyle \Rightarrow x\ge-3.
\displaystyle \therefore \text{Domain}(f)=[-3,\infty).
\displaystyle \text{Since }f\text{ is a square root function, Range}(f)=[0,\infty).
\displaystyle \therefore f:[-3,\infty)\rightarrow[0,\infty).
\displaystyle g(x)=x^2+1\text{ is a polynomial.}
\displaystyle \therefore g:\mathbb R\rightarrow[1,\infty).
\displaystyle \text{Since Range}(g)=[1,\infty)\subseteq\text{Domain}(f)=[-3,\infty),
\displaystyle f\circ g\text{ exists on }\mathbb R.
\displaystyle (f\circ g)(x)=f(g(x))=f(x^2+1)=\sqrt{x^2+1+3}=\sqrt{x^2+4}.
\displaystyle \therefore f\circ g:\mathbb R\rightarrow[2,\infty)\text{ is given by}
\displaystyle (f\circ g)(x)=\sqrt{x^2+4}.
\displaystyle \text{Since Range}(f)=[0,\infty)\subseteq\text{Domain}(g)=\mathbb R,
\displaystyle g\circ f\text{ exists on }[-3,\infty).
\displaystyle (g\circ f)(x)=g(f(x))=(\sqrt{x+3})^2+1=x+3+1=x+4.
\displaystyle \therefore g\circ f:[-3,\infty)\rightarrow[1,\infty)\text{ is given by}
\displaystyle (g\circ f)(x)=x+4.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Let }f\text{ be a real function given by }f(x)=\sqrt{x-2}.\text{ Find each of the following:}
\displaystyle \text{(i) }f\circ f\qquad\text{(ii) }f\circ f\circ f\qquad\text{(iii) }(f\circ f\circ f)(38)\qquad\text{(iv) }f^2.
\displaystyle \text{Also, show that }f\circ f\ne f^2.
\displaystyle \text{Answer:}
\displaystyle f(x)=\sqrt{x-2}.
\displaystyle \text{For the domain of }f,\quad x-2\ge0.
\displaystyle \Rightarrow x\ge2.
\displaystyle \therefore \text{Domain}(f)=[2,\infty).
\displaystyle \text{Since }f(x)\ge0\text{ and every non-negative value is attained,}
\displaystyle \text{Range}(f)=[0,\infty).
\displaystyle \therefore f:[2,\infty)\rightarrow[0,\infty).

\displaystyle \text{(i) To find }f\circ f,\text{ we require }x\in\text{Domain}(f)\text{ and }f(x)\in\text{Domain}(f).
\displaystyle \sqrt{x-2}\ge2.
\displaystyle \Rightarrow x-2\ge4.
\displaystyle \Rightarrow x\ge6.
\displaystyle \therefore \text{Domain}(f\circ f)=[6,\infty).
\displaystyle (f\circ f)(x)=f(f(x))=f(\sqrt{x-2}).
\displaystyle =\sqrt{\sqrt{x-2}-2}.
\displaystyle \therefore f\circ f:[6,\infty)\rightarrow[0,\infty)\text{ is given by}
\displaystyle (f\circ f)(x)=\sqrt{\sqrt{x-2}-2}.
\displaystyle \\

\displaystyle \text{(ii) }f\circ f\circ f=(f\circ f)\circ f.
\displaystyle \text{For its domain, we require }x\in\text{Domain}(f)\text{ and }f(x)\in\text{Domain}(f\circ f).
\displaystyle \sqrt{x-2}\ge6.
\displaystyle \Rightarrow x-2\ge36.
\displaystyle \Rightarrow x\ge38.
\displaystyle \therefore \text{Domain}(f\circ f\circ f)=[38,\infty).
\displaystyle (f\circ f\circ f)(x)=(f\circ f)(f(x)).
\displaystyle =(f\circ f)(\sqrt{x-2}).
\displaystyle =\sqrt{\sqrt{\sqrt{x-2}-2}-2}.
\displaystyle \therefore f\circ f\circ f:[38,\infty)\rightarrow[0,\infty)\text{ is given by}
\displaystyle (f\circ f\circ f)(x)=\sqrt{\sqrt{\sqrt{x-2}-2}-2}.
\displaystyle \\

\displaystyle \text{(iii) }(f\circ f\circ f)(38)=\sqrt{\sqrt{\sqrt{38-2}-2}-2}.
\displaystyle =\sqrt{\sqrt{6-2}-2}.
\displaystyle =\sqrt{2-2}=0.
\displaystyle \therefore (f\circ f\circ f)(38)=0.
\displaystyle \\

\displaystyle \text{(iv) }f^2(x)=f(x)\times f(x).
\displaystyle =\sqrt{x-2}\times\sqrt{x-2}=x-2,\qquad x\in[2,\infty).
\displaystyle \therefore f^2:[2,\infty)\rightarrow[0,\infty)\text{ is given by }f^2(x)=x-2.
\displaystyle \text{Now, at }x=6,
\displaystyle (f\circ f)(6)=\sqrt{\sqrt{6-2}-2}=0,
\displaystyle \text{whereas }f^2(6)=6-2=4.
\displaystyle \therefore f\circ f\ne f^2.
\displaystyle \\

\displaystyle \textbf{Question 12: }f(x)=\begin{cases}1+x,&0\leq x\leq2\\3-x,&2<x\leq3\end{cases}.\quad\text{Find }f\circ f.
\displaystyle \text{Answer:}
\displaystyle f(x)=\begin{cases}1+x,&0\leq x\leq2\\3-x,&2<x\leq3.\end{cases}
\displaystyle \text{The domain of }f\text{ is }[0,3].
\displaystyle \text{For }0\leq x\leq2,\quad 1\leq f(x)\leq3.
\displaystyle \text{For }2<x\leq3,\quad 0\leq f(x)<1.
\displaystyle \therefore \text{Range}(f)=[0,3]=\text{Domain}(f).
\displaystyle \text{Hence, }f\circ f\text{ is defined on }[0,3].
\displaystyle \text{When }0\leq x\leq1,
\displaystyle f(x)=1+x\text{ and }1\leq1+x\leq2.
\displaystyle \therefore f(f(x))=1+(1+x)=x+2.
\displaystyle \text{When }1<x\leq2,
\displaystyle f(x)=1+x\text{ and }2<1+x\leq3.
\displaystyle \therefore f(f(x))=3-(1+x)=2-x.
\displaystyle \text{When }2<x\leq3,
\displaystyle f(x)=3-x\text{ and }0\leq3-x<1.
\displaystyle \therefore f(f(x))=1+(3-x)=4-x.
\displaystyle \therefore (f\circ f)(x)=\begin{cases}x+2,&0\leq x\leq1\\2-x,&1<x\leq2\\4-x,&2<x\leq3.\end{cases}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If }f,g:\mathbb R\rightarrow\mathbb R\text{ are defined by }f(x)=|x|+x
\displaystyle \text{ and }g(x)=|x|-x,\ \forall x\in\mathbb R,\text{ find }f\circ g\text{ and }g\circ f.
\displaystyle \text{Hence, find }(f\circ g)(-3),\ (f\circ g)(5)\text{ and }(g\circ f)(-2).
\displaystyle \text{Answer:}
\displaystyle \text{Given }f(x)=|x|+x\text{ and }g(x)=|x|-x,\ \forall x\in\mathbb R.
\displaystyle \text{Since }f,g:\mathbb R\rightarrow\mathbb R,\text{ both compositions exist.}
\displaystyle (f\circ g)(x)=f(g(x))=|g(x)|+g(x)=||x|-x|+(|x|-x).
\displaystyle \text{If }x\ge0,\text{ then }g(x)=x-x=0.
\displaystyle \therefore (f\circ g)(x)=0.
\displaystyle \text{If }x<0,\text{ then }g(x)=-x-x=-2x>0.
\displaystyle \therefore (f\circ g)(x)=|-2x|+(-2x)=(-2x)+(-2x)=-4x.
\displaystyle \therefore (f\circ g)(x)=\begin{cases}0,&x\ge0\\-4x,&x<0.\end{cases}
\displaystyle \\

\displaystyle (g\circ f)(x)=g(f(x))=|f(x)|-f(x)=||x|+x|-(|x|+x).
\displaystyle \text{If }x\ge0,\text{ then }f(x)=2x\ge0.
\displaystyle \therefore (g\circ f)(x)=2x-2x=0.
\displaystyle \text{If }x<0,\text{ then }f(x)=(-x)+x=0.
\displaystyle \therefore (g\circ f)(x)=0.
\displaystyle \therefore (g\circ f)(x)=0,\ \forall x\in\mathbb R.
\displaystyle \\

\displaystyle (f\circ g)(-3)=-4(-3)=12.
\displaystyle (f\circ g)(5)=0.
\displaystyle (g\circ f)(-2)=0.
\displaystyle \\

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