Class 11: Functions – Exercise 3.3

$\displaystyle \textbf{Question 1: }\text{Find the domain of each of the following real-valued functions of a real}$
$\displaystyle \text{variable:}$
$\displaystyle \text{(i) }f(x)=\frac{1}{x}\qquad \text{(ii) }f(x)=\frac{1}{x-7}\qquad \text{(iii) }f(x)=\frac{3x-2}{x-1}$
$\displaystyle \text{(iv) }f(x)=\frac{2x+1}{x^2-9}\qquad \text{(v) }f(x)=\frac{x^2+2x+1}{x^2-8x+12}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Given }f(x)=\frac{1}{x}$
$\displaystyle \text{The function is defined when }x\neq0.$
$\displaystyle \therefore \text{Domain}(f)=\mathbb{R}-\{0\}.$
$\displaystyle \text{(ii) Given }f(x)=\frac{1}{x-7}$
$\displaystyle \text{The function is defined when }x-7\neq0.$
$\displaystyle \Rightarrow x\neq7$
$\displaystyle \therefore \text{Domain}(f)=\mathbb{R}-\{7\}.$
$\displaystyle \text{(iii) Given }f(x)=\frac{3x-2}{x-1}$
$\displaystyle \text{The function is defined when }x-1\neq0.$
$\displaystyle \Rightarrow x\neq1$
$\displaystyle \therefore \text{Domain}(f)=\mathbb{R}-\{1\}.$
$\displaystyle \text{(iv) Given }f(x)=\frac{2x+1}{x^2-9}=\frac{2x+1}{(x-3)(x+3)}$
$\displaystyle \text{The function is defined when }(x-3)(x+3)\neq0.$
$\displaystyle \Rightarrow x\neq-3,\ 3$
$\displaystyle \therefore \text{Domain}(f)=\mathbb{R}-\{-3,3\}.$
$\displaystyle \text{(v) Given }f(x)=\frac{x^2+2x+1}{x^2-8x+12}$
$\displaystyle =\frac{x^2+2x+1}{(x-6)(x-2)}$
$\displaystyle \text{The function is defined when }(x-6)(x-2)\neq0.$
$\displaystyle \Rightarrow x\neq2,\ 6$
$\displaystyle \therefore \text{Domain}(f)=\mathbb{R}-\{2,6\}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Find the domain of each of the following real-valued functions of a real}$
$\displaystyle \text{variable:}$
$\displaystyle \text{(i) }f(x)=\sqrt{x-2}\qquad \text{(ii) }f(x)=\frac{1}{\sqrt{x^2-1}}\qquad \text{(iii) }f(x)=\sqrt{9-x^2}$
$\displaystyle \text{(iv) }f(x)=\sqrt{\frac{x-2}{3-x}}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Given }f(x)=\sqrt{x-2}$
$\displaystyle \text{For }f(x)\text{ to be real, }x-2\ge0.$
$\displaystyle \Rightarrow x\ge2$
$\displaystyle \therefore \text{Domain}(f)=[2,\infty).$
$\displaystyle \text{(ii) Given }f(x)=\frac{1}{\sqrt{x^2-1}}$
$\displaystyle \text{Since the square root is in the denominator, }x^2-1>0.$
$\displaystyle \Rightarrow (x-1)(x+1)>0$
$\displaystyle \Rightarrow x<-1\text{ or }x>1$
$\displaystyle \therefore \text{Domain}(f)=(-\infty,-1)\cup(1,\infty).$
$\displaystyle \text{(iii) Given }f(x)=\sqrt{9-x^2}$
$\displaystyle \text{For }f(x)\text{ to be real, }9-x^2\ge0.$
$\displaystyle \Rightarrow (3-x)(3+x)\ge0$
$\displaystyle \Rightarrow -3\le x\le3$
$\displaystyle \therefore \text{Domain}(f)=[-3,3].$
$\displaystyle \text{(iv) Given }f(x)=\sqrt{\frac{x-2}{3-x}}$
$\displaystyle \text{For }f(x)\text{ to be real, }\frac{x-2}{3-x}\ge0,\text{ with }x\ne3.$
$\displaystyle \text{Critical points are }x=2\text{ and }x=3.$
$\displaystyle \text{The expression is non-negative only for }2\le x<3.$
$\displaystyle \therefore \text{Domain}(f)=[2,3).$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Find the domain of each of the following real-valued functions of a real}$
$\displaystyle \text{variable:}$
$\displaystyle \text{(i) }f(x)=\frac{ax+b}{bx-a}\qquad \text{(ii) }f(x)=\frac{ax-b}{cx-d}\qquad \text{(iii) }f(x)=\sqrt{x-1}$
$\displaystyle \text{(iv) }f(x)=\sqrt{x-3}\qquad \text{(v) }f(x)=\frac{x-2}{2-x}\qquad \text{(vi) }f(x)=|x-1|$
$\displaystyle \text{(vii) }f(x)=-|x|\qquad \text{(viii) }f(x)=\sqrt{9-x^2}\qquad \text{(ix) }f(x)=\frac{1}{\sqrt{16-x^2}}$
$\displaystyle \text{(x) }f(x)=\sqrt{x^2-16}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Given }f(x)=\frac{ax+b}{bx-a},\text{ where }b\ne0.$
$\displaystyle \text{The function is defined when }bx-a\ne0.$
$\displaystyle \Rightarrow x\ne\frac{a}{b}$
$\displaystyle \therefore \text{Domain}(f)=\mathbb{R}-\left\{\frac{a}{b}\right\}.$
$\displaystyle \text{(ii) Given }f(x)=\frac{ax-b}{cx-d},\text{ where }c\ne0.$
$\displaystyle \text{The function is defined when }cx-d\ne0.$
$\displaystyle \Rightarrow x\ne\frac{d}{c}$
$\displaystyle \therefore \text{Domain}(f)=\mathbb{R}-\left\{\frac{d}{c}\right\}.$
$\displaystyle \text{(iii) Given }f(x)=\sqrt{x-1}$
$\displaystyle \text{For }f(x)\text{ to be real, }x-1\ge0.$
$\displaystyle \Rightarrow x\ge1$
$\displaystyle \therefore \text{Domain}(f)=[1,\infty).$
$\displaystyle \text{(iv) Given }f(x)=\sqrt{x-3}$
$\displaystyle \text{For }f(x)\text{ to be real, }x-3\ge0.$
$\displaystyle \Rightarrow x\ge3$
$\displaystyle \therefore \text{Domain}(f)=[3,\infty).$
$\displaystyle \text{(v) Given }f(x)=\frac{x-2}{2-x}$
$\displaystyle \text{The function is defined when }2-x\ne0.$
$\displaystyle \Rightarrow x\ne2$
$\displaystyle \therefore \text{Domain}(f)=\mathbb{R}-\{2\}.$
$\displaystyle \text{(vi) Given }f(x)=|x-1|$
$\displaystyle \text{The absolute-value expression is defined for every real }x.$
$\displaystyle \therefore \text{Domain}(f)=\mathbb{R}.$
$\displaystyle \text{(vii) Given }f(x)=-|x|$
$\displaystyle \text{The absolute-value expression is defined for every real }x.$
$\displaystyle \therefore \text{Domain}(f)=\mathbb{R}.$
$\displaystyle \text{(viii) Given }f(x)=\sqrt{9-x^2}$
$\displaystyle \text{For }f(x)\text{ to be real, }9-x^2\ge0.$
$\displaystyle \Rightarrow (3-x)(3+x)\ge0$
$\displaystyle \Rightarrow -3\le x\le3$
$\displaystyle \therefore \text{Domain}(f)=[-3,3].$
$\displaystyle \text{(ix) Given }f(x)=\frac{1}{\sqrt{16-x^2}}$
$\displaystyle \text{Since the square root is in the denominator, }16-x^2>0.$
$\displaystyle \Rightarrow (4-x)(4+x)>0$
$\displaystyle \Rightarrow -4<x<4$
$\displaystyle \therefore \text{Domain}(f)=(-4,4).$
$\displaystyle \text{(x) Given }f(x)=\sqrt{x^2-16}$
$\displaystyle \text{For }f(x)\text{ to be real, }x^2-16\ge0.$
$\displaystyle \Rightarrow (x-4)(x+4)\ge0$
$\displaystyle \Rightarrow x\le-4\text{ or }x\ge4$
$\displaystyle \therefore \text{Domain}(f)=(-\infty,-4]\cup[4,\infty).$
$\displaystyle \\$