Relations and Functions — Exercise 2.4

If $f(x) = x^2$, find $\dfrac{f(1.1) - f(1)}{1.1 - 1}$.

Answer

Given,

$f(x) = x^2$

⇒ $f(1.1) = (1.1)^2 = 1.21$ and $f(1) = 1^2 = 1$

$\therefore \dfrac{f(1.1) - f(1)}{1.1 - 1} = \dfrac{1.21 - 1}{1.1 - 1} \\[1em] = \dfrac{0.21}{0.1} \\[1em] = 2.1$

Hence, $\bm{\dfrac{f(1.1) - f(1)}{1.1 - 1} = 2.1}$.

A function is defined by $f(x) = \dfrac{3x^2 + 2x - 1}{x + 1}$, $x \in \textbf{R}$, $x \ne -1$. Find the value of $f(-3) + 1$.

Answer

Given,

$f(x) = \dfrac{3x^2 + 2x - 1}{x + 1}$, $x \ne -1$

As $-3 \in D_f$, $f(-3)$ exists.

Substituting $x = -3$, we get :

$f(-3) = \dfrac{3(-3)^2 + 2(-3) - 1}{-3 + 1}$

⇒ $f(-3) = \dfrac{27 - 6 - 1}{-2}$

⇒ $f(-3) = \dfrac{20}{-2}$

⇒ $f(-3) = -10$

∴ $f(-3) + 1 = -10 + 1 = -9$

Hence, $\bm{f(-3) + 1 = -9}$.

Is the relation $f$ defined by

$f(x) = \begin{cases} x^2, & 0 \le x \le 3 \\[1em] 2x, & 3 \le x \le 5 \end{cases}$

a function? Justify your answer.

Answer

Given,

$f(x) = \begin{cases} x^2, & 0 \le x \le 3 \\[1em] 2x, & 3 \le x \le 5 \end{cases}$

The value $x = 3$ lies in both the intervals $0 \le x \le 3$ and $3 \le x \le 5$, so both the rules apply to $x = 3$.

By the first rule,

$f(3) = 3^2 = 9$

By the second rule,

$f(3) = 2 \times 3 = 6$

Thus the element 3 of the domain of $f$ has two different images 9 and 6.

Hence, the given relation $\bm{f}$ is not a function, because the element 3 of its domain has two different images 9 and 6.

A real function $f$ is defined by $f(x) = 3x - 2$, $x \in \textbf{R}$, find the value of $f(-3) + f(0) \times f(7)$.

Answer

Given,

$f(x) = 3x - 2$, $D_f = \textbf{R}$

Substituting different values of $x$, we get :

⇒ $f(-3) = 3(-3) - 2 = -9 - 2 = -11$,

⇒ $f(0) = 3(0) - 2 = 0 - 2 = -2$,

⇒ $f(7) = 3(7) - 2 = 21 - 2 = 19$.

∴ $f(-3) + f(0) \times f(7) = -11 + (-2) \times 19$

⇒ $f(-3) + f(0) \times f(7) = -11 - 38$

⇒ $f(-3) + f(0) \times f(7) = -49$

Hence, $\bm{f(-3) + f(0) \times f(7) = -49}$.

Given $f(x) = x^3 - 1$, find $x$ if $f(x) = 215$.

Answer

Given,

$f(x) = x^3 - 1$ and $f(x) = 215$

⇒ $x^3 - 1 = 215$

⇒ $x^3 = 215 + 1$

⇒ $x^3 = 216$

⇒ $x^3 = 6^3$

⇒ $x = 6$

Hence, $\bm{x = 6}$.

If $f(x) = 2x^2 + 1$ and domain of $f = {-2, -1, 0, 1, 3}$, find

(i) range of $f$

(ii) $f(3) \times f(-2)$

(iii) $x$ if $f(x) = 9$.

Answer

Given,

$f(x) = 2x^2 + 1$ and $D_f = {-2, -1, 0, 1, 3}$

(i) Substituting each value of $x$ in $f(x) = 2x^2 + 1$, we get :

$f(-2) = 2(-2)^2 + 1 = 8 + 1 = 9$,

$f(-1) = 2(-1)^2 + 1 = 2 + 1 = 3$,

$f(0) = 2(0)^2 + 1 = 0 + 1 = 1$,

$f(1) = 2(1)^2 + 1 = 2 + 1 = 3$,

$f(3) = 2(3)^2 + 1 = 18 + 1 = 19$.

$R_f$ is the set of all these images.

Hence, range of $\bm{f = {1, 3, 9, 19}}$.

(ii) From part (i), $f(3) = 19$ and $f(-2) = 9$.

∴ $f(3) \times f(-2) = 19 \times 9 = 171$

Hence, $\bm{f(3) \times f(-2) = 171}$.

(iii) Given, $f(x) = 9$

⇒ $2x^2 + 1 = 9$

⇒ $2x^2 = 9 - 1$

⇒ $2x^2 = 8$

⇒ $x^2 = 4$

⇒ $x = \pm 2$

Out of 2 and $-2$, only $-2$ belongs to $D_f$.

Hence, $\bm{x = -2}$.

A function '$f$' is defined by $f(x) = x^2 + 3$, $x \in \textbf{N}$ and $x \le 5$.

(i) Find the range of $f$.

(ii) Find $f(2) \times f(5)$.

(iii) Does $f(-3)$ exist?

(iv) Find $x$ when $f(x) = 7$.

Answer

Given,

$f(x) = x^2 + 3$, $x \in \textbf{N}$ and $x \le 5$

The natural numbers not exceeding 5 are 1, 2, 3, 4 and 5.

∴ $D_f = {1, 2, 3, 4, 5}$

(i) Substituting each value of $x$ in $f(x) = x^2 + 3$, we get :

$f(1) = 1^2 + 3 = 1 + 3 = 4$,

$f(2) = 2^2 + 3 = 4 + 3 = 7$,

$f(3) = 3^2 + 3 = 9 + 3 = 12$,

$f(4) = 4^2 + 3 = 16 + 3 = 19$,

$f(5) = 5^2 + 3 = 25 + 3 = 28$.

$R_f$ is the set of all these images.

Hence, range of $\bm{f = {4, 7, 12, 19, 28}}$.

(ii) From part (i), $f(2) = 7$ and $f(5) = 28$.

∴ $f(2) \times f(5) = 7 \times 28 = 196$

Hence, $\bm{f(2) \times f(5) = 196}$.

(iii) $D_f = {1, 2, 3, 4, 5}$ and $-3$ does not belong to it, as $-3$ is not a natural number.

Hence, $\bm{f(-3)}$ does not exist.

(iv) Given, $f(x) = 7$

⇒ $x^2 + 3 = 7$

⇒ $x^2 = 7 - 3$

⇒ $x^2 = 4$

⇒ $x = \pm 2$

Out of 2 and $-2$, only 2 belongs to $D_f$.

Hence, $\bm{x = 2}$.

Find the domain of the following real functions:

(i) $f(x) = \dfrac{x^2 - 4}{x + 2}$

(ii) $f(x) = \dfrac{3x}{28 - x}$

(iii) $f(x) = \dfrac{x}{x^2 + 3x + 2}$

(iv) $f(x) = \dfrac{x^3 - x + 3}{x^2 - 1}$.

Answer

(i) $f(x) = \dfrac{x^2 - 4}{x + 2}$

For $D_f$, $f(x)$ must be a real number

⇒ $\dfrac{x^2 - 4}{x + 2}$ must be a real number

⇒ $x + 2 \ne 0$

⇒ $x \ne -2$

⇒ $D_f$ = set of all real numbers except $-2$ i.e. $D_f = \textbf{R} - {-2}$

Hence, domain of $\bm{f = \textbf{R} - {-2}}$.

(ii) $f(x) = \dfrac{3x}{28 - x}$

For $D_f$, $f(x)$ must be a real number

⇒ $\dfrac{3x}{28 - x}$ must be a real number

⇒ $28 - x \ne 0$

⇒ $x \ne 28$

⇒ $D_f$ = set of all real numbers except 28 i.e. $D_f = \textbf{R} - {28}$

Hence, domain of $\bm{f = \textbf{R} - {28}}$.

(iii) $f(x) = \dfrac{x}{x^2 + 3x + 2}$

For $D_f$, $f(x)$ must be a real number

⇒ $\dfrac{x}{x^2 + 3x + 2}$ must be a real number

⇒ $x^2 + 3x + 2 \ne 0$

⇒ $x^2 + 2x + x + 2 \ne 0$

⇒ $x(x + 2) + 1(x + 2) \ne 0$

⇒ $(x + 2)(x + 1) \ne 0$

⇒ $x \ne -2, -1$

⇒ $D_f$ = set of all real numbers except $-2$ and $-1$ i.e. $D_f = \textbf{R} - {-1, -2}$

Hence, domain of $\bm{f = \textbf{R} - {-1, -2}}$.

(iv) $f(x) = \dfrac{x^3 - x + 3}{x^2 - 1}$

For $D_f$, $f(x)$ must be a real number

⇒ $\dfrac{x^3 - x + 3}{x^2 - 1}$ must be a real number

⇒ $x^2 - 1 \ne 0$

⇒ $(x - 1)(x + 1) \ne 0$

⇒ $x \ne 1, -1$

⇒ $D_f$ = set of all real numbers except $-1$ and 1 i.e. $D_f = \textbf{R} - {-1, 1}$

Hence, domain of $\bm{f = \textbf{R} - {-1, 1}}$.

Find the range of the following functions:

(i) $f(x) = x^2 + 2$, $x \in \textbf{R}$

(ii) $f(x) = 3 - 2x$, $x \in \textbf{R}$, $x \ge 1$.

Answer

(i) $f(x) = x^2 + 2$, $x \in \textbf{R}$

For $R_f$, let $y = f(x) = x^2 + 2$, $x \in \textbf{R}$.

We know that for all $x \in \textbf{R}$, $x^2 \ge 0$

⇒ $x^2 + 2 \ge 0 + 2$

⇒ $y \ge 2$

⇒ $R_f = [2, \infty)$

Hence, range of $\bm{f = [2, \infty)}$.

(ii) $f(x) = 3 - 2x$, $x \in \textbf{R}$, $x \ge 1$

For $R_f$, let $y = f(x) = 3 - 2x$, $x \in \textbf{R}$, $x \ge 1$.

Now, $x \ge 1$

⇒ $2x \ge 2$

⇒ $-2x \le -2$

⇒ $3 - 2x \le 3 - 2$

⇒ $y \le 1$

⇒ $R_f = (-\infty, 1]$

Hence, range of $\bm{f = (-\infty, 1]}$.

Find the domain and the range of the function $f(x) = 2x^2 + 1$. Also find $f(-2)$ and the numbers which are associated with the number 51 in its range.

Answer

Given,

$f(x) = 2x^2 + 1$

For $D_f$, $f(x)$ must be a real number

⇒ $2x^2 + 1$ must be a real number, which is a real number for every $x \in \textbf{R}$

⇒ $D_f = \textbf{R}$

For $R_f$, let $y = f(x) = 2x^2 + 1$ ...........(1)

We know that for all $x \in \textbf{R}$, $x^2 \ge 0$

⇒ $2x^2 \ge 0$

⇒ $2x^2 + 1 \ge 0 + 1$

⇒ $y \ge 1$

⇒ $R_f = [1, \infty)$

Further, as $-2 \in D_f$, $f(-2)$ exists and $f(-2) = 2(-2)^2 + 1 = 8 + 1 = 9$.

As $51 \in R_f$, on putting $y = 51$ in equation (1), we get :

⇒ $2x^2 + 1 = 51$

⇒ $2x^2 = 51 - 1$

⇒ $2x^2 = 50$

⇒ $x^2 = 25$

⇒ $x = \pm 5$

Therefore, 5 and $-5$ are the numbers (in $D_f$) which are associated with the number 51 in $R_f$.

Hence, domain of $\bm{f = \textbf{R}}$, range of $\bm{f = [1, \infty)}$, $\bm{f(-2) = 9}$ and the numbers associated with 51 are 5 and $\bm{-5}$.

Find the domain and the range of the following functions:

$f(x) = \sqrt{x + 2}$

Answer

Given,

$f(x) = \sqrt{x + 2}$

For $D_f$, $f(x)$ must be a real number

⇒ $\sqrt{x + 2}$ must be a real number

⇒ $x + 2 \ge 0$

⇒ $x \ge -2$

⇒ $D_f = [-2, \infty)$

For $R_f$, let $y = f(x) = \sqrt{x + 2}$.

As $x \ge -2$, $x + 2 \ge 0$

⇒ $\sqrt{x + 2} \ge 0$

⇒ $y \ge 0$

⇒ $R_f = [0, \infty)$

Hence, domain of $\bm{f = [-2, \infty)}$ and range of $\bm{f = [0, \infty)}$.

Find the domain and the range of the following functions:

$f(x) = \sqrt{3 - 2x}$

Answer

Given,

$f(x) = \sqrt{3 - 2x}$

For $D_f$, $f(x)$ must be a real number

⇒ $\sqrt{3 - 2x}$ must be a real number

⇒ $3 - 2x \ge 0$

⇒ $3 \ge 2x$

⇒ $x \le \dfrac{3}{2}$

⇒ $D_f = \left(-\infty, \dfrac{3}{2}\right]$

For $R_f$, let $y = f(x) = \sqrt{3 - 2x}$.

As $x \le \dfrac{3}{2}$, $2x \le 3$ i.e. $3 - 2x \ge 0$

⇒ $\sqrt{3 - 2x} \ge 0$

⇒ $y \ge 0$

⇒ $R_f = [0, \infty)$

Hence, domain of $\bm{f = \left(-\infty, \dfrac{3}{2}\right]}$ and range of $\bm{f = [0, \infty)}$.

Find the domain and the range of the following functions:

$f(x) = \dfrac{1}{\sqrt{x - 5}}$.

Answer

Given,

$f(x) = \dfrac{1}{\sqrt{x - 5}}$

For $D_f$, $f(x)$ must be a real number

⇒ $\dfrac{1}{\sqrt{x - 5}}$ must be a real number

⇒ $x - 5 \gt 0$

⇒ $x \gt 5$

⇒ $D_f = (5, \infty)$

For $R_f$, let $y = f(x) = \dfrac{1}{\sqrt{x - 5}}$.

As $x \gt 5$, $0 \lt x - 5$

⇒ $\sqrt{x - 5} \gt 0$

⇒ $\dfrac{1}{\sqrt{x - 5}} \gt 0$ $\left(\because \dfrac{1}{a} \gt 0 \text{ if and only if } a \gt 0\right)$

⇒ $y \gt 0$

⇒ $R_f = (0, \infty)$

Hence, domain of $\bm{f = (5, \infty)}$ and range of $\bm{f = (0, \infty)}$.

Find the domain and the range of the function :

$f(x) = \dfrac{4 - x}{x - 4}$

Answer

Given,

$f(x) = \dfrac{4 - x}{x - 4}$

For $D_f$, $f(x)$ must be a real number

⇒ $\dfrac{4 - x}{x - 4}$ must be a real number

⇒ $x - 4 \ne 0$

⇒ $x \ne 4$

⇒ $D_f$ = set of all real numbers except 4 i.e. $D_f = \textbf{R} - {4}$

For $R_f$, let $y = \dfrac{4 - x}{x - 4}$.

For every $x \ne 4$,

⇒ $y = \dfrac{-(x - 4)}{x - 4}$

⇒ $y = -1$

So $f(x)$ takes only the value $-1$.

⇒ $R_f = {-1}$

Hence, domain of $\bm{f = \textbf{R} - {4}}$ and range of $\bm{f = {-1}}$.

Find the domain and the range of the function :

$f(x) = \dfrac{x^2 - 9}{x - 3}$

Answer

Given,

$f(x) = \dfrac{x^2 - 9}{x - 3}$

For $D_f$, $f(x)$ must be a real number

⇒ $\dfrac{x^2 - 9}{x - 3}$ must be a real number

⇒ $x - 3 \ne 0$

⇒ $x \ne 3$

⇒ $D_f$ = set of all real numbers except 3 i.e. $D_f = \textbf{R} - {3}$

For $R_f$, let $y = \dfrac{x^2 - 9}{x - 3}$.

For every $x \ne 3$,

⇒ $y = \dfrac{(x - 3)(x + 3)}{x - 3}$

⇒ $y = x + 3$

As $x$ takes every real value except 3, $x + 3$ takes every real value except $3 + 3 = 6$.

⇒ $R_f = \textbf{R} - {6}$

Hence, domain of $\bm{f = \textbf{R} - {3}}$ and range of $\bm{f = \textbf{R} - {6}}$.

Find the domain and the range of the function :

$f(x) = \dfrac{x + 1}{2x + 1}$

Answer

Given,

$f(x) = \dfrac{x + 1}{2x + 1}$

For $D_f$, $f(x)$ must be a real number

⇒ $\dfrac{x + 1}{2x + 1}$ must be a real number

⇒ $2x + 1 \ne 0$

⇒ $x \ne -\dfrac{1}{2}$

⇒ $D_f$ = set of all real numbers except $-\dfrac{1}{2}$ i.e. $D_f = \textbf{R} - \bigg\lbrace-\dfrac{1}{2}\bigg\rbrace$

For $R_f$, let $y = \dfrac{x + 1}{2x + 1}$

⇒ $y(2x + 1) = x + 1$

⇒ $2xy + y = x + 1$

⇒ $2xy - x = 1 - y$

⇒ $(2y - 1)x = 1 - y$

⇒ $x = \dfrac{1 - y}{2y - 1}$ but $x \in \textbf{R}$

⇒ $\dfrac{1 - y}{2y - 1}$ must be a real number

⇒ $2y - 1 \ne 0$

⇒ $y \ne \dfrac{1}{2}$

⇒ $R_f$ = set of all real numbers except $\dfrac{1}{2}$ i.e. $R_f = \textbf{R} - \bigg\lbrace\dfrac{1}{2}\bigg\rbrace$

Hence, domain of $\bm{f = \textbf{R} - \bigg\lbrace-\dfrac{1}{2}\bigg\rbrace}$ and range of $\bm{f = \textbf{R} - \bigg\lbrace\dfrac{1}{2}\bigg\rbrace}$.

Find the domain and the range of the function :

$f(x) = \dfrac{4 + x}{4 - x}$

Answer

Given,

$f(x) = \dfrac{4 + x}{4 - x}$

For $D_f$, $f(x)$ must be a real number

⇒ $\dfrac{4 + x}{4 - x}$ must be a real number

⇒ $4 - x \ne 0$

⇒ $x \ne 4$

⇒ $D_f$ = set of all real numbers except 4 i.e. $D_f = \textbf{R} - {4}$

For $R_f$, let $y = \dfrac{4 + x}{4 - x}$

⇒ $y(4 - x) = 4 + x$

⇒ $4y - xy = 4 + x$

⇒ $4y - 4 = x + xy$

⇒ $(1 + y)x = 4y - 4$

⇒ $x = \dfrac{4y - 4}{y + 1}$ but $x \in \textbf{R}$

⇒ $\dfrac{4y - 4}{y + 1}$ must be a real number

⇒ $y + 1 \ne 0$

⇒ $y \ne -1$

⇒ $R_f$ = set of all real numbers except $-1$ i.e. $R_f = \textbf{R} - {-1}$

Hence, domain of $\bm{f = \textbf{R} - {4}}$ and range of $\bm{f = \textbf{R} - {-1}}$.

Find the domain and the range of the function :

$f(x) = \dfrac{1}{4 - x^2}$.

Answer

Given,

$f(x) = \dfrac{1}{4 - x^2}$

For $D_f$, $f(x)$ must be a real number

⇒ $\dfrac{1}{4 - x^2}$ must be a real number

⇒ $4 - x^2 \ne 0$

⇒ $(2 - x)(2 + x) \ne 0$

⇒ $x \ne \pm 2$

⇒ $D_f$ = set of all real numbers except $-2$ and 2 i.e. $D_f = \textbf{R} - {-2, 2}$

For $R_f$, let $y = \dfrac{1}{4 - x^2}$

⇒ $4 - x^2 = \dfrac{1}{y}$, $y \ne 0$

⇒ $x^2 = 4 - \dfrac{1}{y}$, $y \ne 0$

As $x^2 \ge 0$ for all $x \in D_f$

⇒ $4 - \dfrac{1}{y} \ge 0$, $y \ne 0$ but y^2 > 0

(Multiplying both the sides by $y^2$, a positive real number)

⇒ $y^2\left(4 - \dfrac{1}{y}\right) \ge 0$, $y \ne 0$

⇒ $y(4y - 1) \ge 0$, $y \ne 0$

⇒ $4y\left(y - \dfrac{1}{4}\right) \ge 0$, $y \ne 0$

⇒ either $y \le 0$ or $y \ge \dfrac{1}{4}$ but $y \ne 0$

⇒ $y \lt 0$ or $y \ge \dfrac{1}{4}$

⇒ $R_f = (-\infty, 0) \cup \left[\dfrac{1}{4}, \infty\right)$

Hence, domain of $\bm{f = \textbf{R} - {-2, 2}}$ and range of $\bm{f = (-\infty, 0) \cup \left[\dfrac{1}{4}, \infty\right)}$.

Find the domain and the range of the following functions:

$f(x) = \sqrt{16 - x^2}$

Answer

Given,

$f(x) = \sqrt{16 - x^2}$

For $D_f$, $f(x)$ must be a real number

⇒ $\sqrt{16 - x^2}$ must be a real number

⇒ $16 - x^2 \ge 0$

⇒ $-(x^2 - 16) \ge 0$

⇒ $x^2 - 16 \le 0$

⇒ $(x + 4)(x - 4) \le 0$

⇒ $-4 \le x \le 4$

⇒ $D_f = [-4, 4]$

For $R_f$, let $y = \sqrt{16 - x^2}$ ...........(1)

As the square root of a real number is always non-negative, $y \ge 0$.

On squaring equation (1), we get :

⇒ $y^2 = 16 - x^2$

⇒ $x^2 = 16 - y^2$ but $x^2 \ge 0$ for all $x \in D_f$

⇒ $16 - y^2 \ge 0$

⇒ $y^2 - 16 \le 0$

⇒ $(y + 4)(y - 4) \le 0$

⇒ $-4 \le y \le 4$ but $y \ge 0$

⇒ $R_f = [0, 4]$

Hence, domain of $\bm{f = [-4, 4]}$ and range of $\bm{f = [0, 4]}$.

Find the domain and the range of the following functions:

$f(x) = \sqrt{x^2 - 9}$

Answer

Given,

$f(x) = \sqrt{x^2 - 9}$

For $D_f$, $f(x)$ must be a real number

⇒ $\sqrt{x^2 - 9}$ must be a real number

⇒ $x^2 - 9 \ge 0$

⇒ $(x + 3)(x - 3) \ge 0$

⇒ either $x \le -3$ or $x \ge 3$

⇒ $D_f = (-\infty, -3] \cup [3, \infty)$

For $R_f$, let $y = \sqrt{x^2 - 9}$ ...........(1)

As the square root of a real number is always non-negative, $y \ge 0$.

On squaring equation (1), we get :

⇒ $y^2 = x^2 - 9$

⇒ $x^2 = y^2 + 9$ but $x^2 \ge 0$ for all $x \in D_f$

⇒ $y^2 + 9 \ge 0$

⇒ $y^2 \ge -9$, which is true for all $y \in \textbf{R}$. Also $y \ge 0$

⇒ $R_f = [0, \infty)$

Hence, domain of $\bm{f = (-\infty, -3] \cup [3, \infty)}$ and range of $\bm{f = [0, \infty)}$.

Find the domain and the range of the following functions:

$f(x) = \dfrac{1}{\sqrt{4 - x^2}}$.

Answer

Given,

$f(x) = \dfrac{1}{\sqrt{4 - x^2}}$

For $D_f$, $f(x)$ must be a real number

⇒ $\dfrac{1}{\sqrt{4 - x^2}}$ must be a real number

⇒ $4 - x^2 \gt 0$

⇒ $-(x^2 - 4) \gt 0$

⇒ $x^2 - 4 \lt 0$

⇒ $(x + 2)(x - 2) \lt 0$

⇒ $-2 \lt x \lt 2$

⇒ $D_f = (-2, 2)$

For $R_f$, let $y = \dfrac{1}{\sqrt{4 - x^2}}$, $y \ne 0$ ...........(1)

Also, as the square root of a real number is always non-negative, $y \gt 0$.

On squaring equation (1), we get :

⇒ $y^2 = \dfrac{1}{4 - x^2}$

⇒ $4 - x^2 = \dfrac{1}{y^2}$

⇒ $x^2 = 4 - \dfrac{1}{y^2}$

But $x^2 \ge 0$ for all $x \in D_f$

⇒ $4 - \dfrac{1}{y^2} \ge 0$ but y^2 > 0

(Multiplying both the sides by $y^2$, a positive real number)

⇒ $4y^2 - 1 \ge 0$

⇒ $y^2 - \dfrac{1}{4} \ge 0$

⇒ $\left(y + \dfrac{1}{2}\right)\left(y - \dfrac{1}{2}\right) \ge 0$

⇒ either $y \le -\dfrac{1}{2}$ or $y \ge \dfrac{1}{2}$ but $y \gt 0$

⇒ $y \ge \dfrac{1}{2}$

⇒ $R_f = \left[\dfrac{1}{2}, \infty\right)$

Hence, domain of $\bm{f = (-2, 2)}$ and range of $\bm{f = \left[\dfrac{1}{2}, \infty\right)}$.

Find the domain and the range of the following functions:

(i) $f(x) = |x - 3|$

(ii) $f(x) = 3 - |x - 2|$.

Answer

(i) $f(x) = |x - 3|$

For $D_f$, $f(x)$ must be a real number

⇒ $|x - 3|$ must be a real number, which is a real number for every $x \in \textbf{R}$

⇒ $D_f = \textbf{R}$

For $R_f$, let $y = f(x) = |x - 3|$.

Since $|x| \ge 0$ for all $x \in \textbf{R}$

⇒ $|x - 3| \ge 0$ for all $x \in \textbf{R}$

⇒ $y \ge 0$

⇒ $R_f = [0, \infty)$

Hence, domain of $\bm{f = \textbf{R}}$ and range of $\bm{f = [0, \infty)}$.

(ii) $f(x) = 3 - |x - 2|$

For $D_f$, $f(x)$ must be a real number

⇒ $3 - |x - 2|$ must be a real number, which is a real number for every $x \in \textbf{R}$

⇒ $D_f = \textbf{R}$

For $R_f$, let $y = f(x) = 3 - |x - 2|$.

Since $|x| \ge 0$ for all $x \in \textbf{R}$