# Examples on Domains and Ranges of Functions Set 2

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Example- 10

Find the domains of the following functions:

 (a) f\left( x \right) =\begin{align} \frac{1}{{\sqrt {{x^2} + 3x + 2} }}\end{align} (b) f\left( x \right) =\begin{align} \frac{1}{{{x^2} + 3x + 3}}\end{align} (c) f\left( x \right) = \begin{align}\frac{1}{{\sqrt {2{x^2} + 5x + 2} }}\end{align} (d) f\left( x \right) =\begin{align} \frac{1}{{{{\left[ x \right]}^2} - 3\left[ x \right] + 2}}\end{align} (e) f\left( x \right) = \begin{align}\frac{1}{{2{{\left[ x \right]}^2} + 5\left[ x \right] + 2}}\end{align} (f) $$f\left( x \right) = \sqrt {{{\left[ x \right]}^2} - 3\left[ x \right] + 2}$$ (g) f\left( x \right) = \begin{align}\frac{1}{{{{\left\{ x \right\}}^2} - 0.25}}\end{align} (h) f\left( x \right) = \begin{align}\frac{1}{x} + \frac{1}{{\left\{ x \right\}}} + \frac{1}{{\left[ x \right]}}\end{align}

Solution:

(a)

\begin{align}f\left( x \right) = \frac{1}{{\sqrt {{x^2} + 3x + 2} }} \Rightarrow \quad & {x^2} + 3x + 2 > 0 \Rightarrow \left( {x + 1} \right)\left( {x + 2} \right) > 0\\ \Rightarrow \quad & x < - 2\,\,{\text{or}}\,\,x > - 1\,\,\,\, \Rightarrow \,\,D = \left( { - \infty ,\,\, - 2} \right) \cup \left( { - 1,\infty } \right)\end{align}

(b)

f\left( x \right) = \begin{align}\frac{1}{{{x^2} + 3x + 3}}\end{align} \Rightarrow {x^2} + 3x + 3 \ne 0

Notice that $${x^2} + 3x + 3$$ can be written as  $${\left( {x + \frac{3}{2}} \right)^2} + \frac{3}{4}$$

which is never 0 $$\Rightarrow D = \mathbb{R}$$.

(c)

\begin{align}f\left( x \right) = \frac{1}{{\sqrt {2{x^2} + 5x + 2} }} \Rightarrow \quad & 2{x^2} + 5x + 2 > 0 \Rightarrow \left( {2x + 1} \right)\left( {x + 2} \right) > 0\\ \Rightarrow \quad & x < - 2\,\,{\rm{or}}\,\,x > - \frac{1}{2} \Rightarrow D = \left( { - \infty , - 2} \right) \cup \left( { - \frac{1}{2},\infty } \right)\end{align}

(d)

\begin{align}f\left( x \right) = \frac{1}{{{{\left[ x \right]}^2} - 3\left[ x \right] + 2}} \Rightarrow \quad & {\left[ x \right]^2} - 3\left[ x \right] + 2 \ne 0 \Rightarrow \left( {\left[ x \right] - 1} \right)\left( {\left[ x \right] - 2} \right) \ne 0 \\ \Rightarrow \quad & \left[ x \right] \ne 1\,\,{\text{or}}\,\,2 \Rightarrow x\not \in \,\,\left[ {1,2} \right)\,\,{\text{or}}\,\left[ {2,3} \right)\, \hfill \\ \Rightarrow \quad & x\not \in \left[ {1,3} \right) \qquad \left\{ {\left[ {1,\,\,2} \right) \cup \left[ {2,3} \right)\,\,{\text{is}}\,{\text{equal}}\,{\text{to}}\,\left[ {1,3} \right)\,} \right\} \hfill \\\Rightarrow \quad & D = \mathbb{R}\backslash \left[ {1,\,\,3} \right) \hfill \\ \end{align}

(e)

\begin{align}f\left( x \right) = \frac{1}{{2{{\left[ x \right]}^2} + 5\left[ x \right] + 2}} \Rightarrow \quad & 2{\left[ x \right]^2} + 5\left[ x \right] + 2 \ne 0 \Rightarrow \left( {2\left[ x \right] + 1} \right)\left( {\left[ x \right] + 2} \right) \ne 0\\ \Rightarrow \quad & \left[ x \right] \ne - \frac{1}{2}\,\,{\rm{or}}\,\, - 2\end{align}

Now obviously, [x] can never equal \begin{align} - \frac{1}{2}\end{align}. Therefore, we require

\begin{align}\left[ x \right] \ne - 2\,\, \Rightarrow \quad & x\not \in \left[ { - 2,\,\, - 1} \right)\\ \Rightarrow \quad & D = \mathbb{R}\backslash \left[ { - 2,\,\, - 1} \right)\end{align}

(f)

$$f\left( x \right) = \sqrt {{{\left[ x \right]}^2} - 3\left[ x \right] + 2} \Rightarrow {\left[ x \right]^2} - 3\left[ x \right] + 2 \ge 0 \Rightarrow \left( {\left[ x \right] - 1} \right)\left( {\left[ x \right] - 2} \right) \ge 0$$

\begin{align} f\left( x \right) = \sqrt {{{\left[ x \right]}^2} - 3\left[ x \right] + 2} \;\;\;\; \Rightarrow \quad & {\left[ x \right]^2} - 3\left[ x \right] + 2 \ge 0 \Rightarrow \left( {\left[ x \right] - 1} \right)\left( {\left[ x \right] - 2} \right) \ge 0\\ \Rightarrow \quad & \left[ x \right] \le 1\,\,{\rm{or}}\,\,\left[ x \right] \ge 2\end{align}

Notice that any value of x will satisfy either of the two constraints above. (These two constraints all the integers!).  $$\Rightarrow D = \mathbb{R}$$

(g)

f\left( x \right) =\begin{align} \frac{1}{{{{\left\{ x \right\}}^2} - 0.25}}\end{align} \Rightarrow {\left\{ x \right\}^2} - 0.25 \ne 0 \Rightarrow \left\{ x \right\} \ne 0.5,\, - 0.5

Observe that {x} can never equal –0.5; hence we require that

$\left\{ x \right\} \ne 0.5\;\;\; \Rightarrow x \ne n + 0.5\;;n \in \mathbb{Z}\;\;\; \Rightarrow D = \mathbb{R}\backslash \left\{ {n + 0.5} \right\};n \in \mathbb{Z}$

(h)

\begin{align}f\left( x \right) = \frac{1}{x} + \frac{1}{{\left\{ x \right\}}} + \frac{1}{{\left[ x \right]}} \Rightarrow \quad & x \ne 0\,\,\,{\rm{and}}\left\{ x \right\} \ne 0\,\,{\rm{and}}\,\left[ x \right] \ne 0\\ \Rightarrow \quad & x \ne 0\,\,{\text{and}}\,\,x\not \in \mathbb{Z}\,\,{\text{and}}\,\,x\not \in \left[ {0,\,\,1} \right) \hfill \\ \Rightarrow \quad & D = \mathbb{R}\backslash \mathbb{Z}\backslash \left( {0,\,\,1} \right) \hfill \\ \end{align}

Example- 11

Find the domains of the following functions:

 (a) $$f\left( x \right) = \sqrt {{x^2} + 3x + 1}$$ (b) $$f(x) = \sqrt {{x^2} - 1} + \frac{1}{{\sqrt {{x^2} - 5x + 4} }}$$ (c) $$f\left( x \right) = \sqrt {x - 2} + \sqrt {4 - x} + \sqrt {9 - {x^2}}$$ (d) $$f\left( x \right) = \sqrt {4 - {x^2}} + \sqrt {{x^2} - 4x + 3}$$ (e) f\left( x \right) =\begin{align} \frac{1}{{\sqrt { - \left( {{x^2} + 3x + 2} \right)} }}\end{align} (f) f\left( x \right) = \begin{align}\frac{1}{{\sqrt { - \left( {{x^2} + 3x + 2} \right)} }} + \sqrt { - \left( {{x^2} - 3x + 2} \right)}\end{align} (g) $$f\left( x \right) = \sqrt {6{x^2} + 5x + 1}$$ (h) f\left( x \right) =\begin{align} \frac{1}{{\sqrt {100 - {x^2}} }} + \sqrt {2 - {x^2}}\end{align}

Solution:

(a) $$f\left( x \right) = \sqrt {{x^2} + 3x + 1} \,\,\, \Rightarrow \,\,\,{x^2} + 3x + 1 \ge 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...{\rm{ }}\left( i \right)$$

The roots of {x^2} + 3x + 1\,\,{\text{are}}\,\,\,\begin{align}\frac{{ - 3 \pm \sqrt 5 }}{2}\end{align}

Hence, the solution to the inequality (i) is

$x \le \frac{{ - 3 - \sqrt 5 }}{2}\;{\rm{or }}x \ge \frac{{ - 3 + \sqrt 5 }}{2} \;\;\; \Rightarrow D = \left( { - \infty ,\frac{{ - 3 - \sqrt 5 \;}}{2}} \right] \cup \left[ {\frac{{ - 3 + \sqrt 5 \;}}{2},\infty } \right)$

(b) Here we can see that f(x) is composed of two separate parts. The domain for f(x) should be such that both parts of f(x) remain defined (real). In all such cases, we find the domain for both parts(or there might also be more than 2 parts) separately. The final domain is the interval(s) common to the domains of all the parts.

Applying this concept here,

$$\Rightarrow$$    $${x^2} - 1 \ge 0$$    $$\Rightarrow$$    $${D_1}$$= $$( - \infty ,\,\, - 1]\,\, \cup \,\,[1,\,\,\infty )$$

$$\Rightarrow$$      $${x^2} - 5x + 4 > 0$$ $$\Rightarrow$$ $${D_2}$$ = $$( - \infty ,\,\,1)\, \cup \,(4,\,\,\infty )$$

The domain for f (x),  \begin{align}&D = {D_1} \cap {D_2}\\& D = ( - \infty ,\, - 1]\,\, \cup \,\,4,\infty ) \end{align}

(c) $$f\left( x \right) = \sqrt {x - 2} + \sqrt {4 - x} + \sqrt {9 - {x^2}} \;\Rightarrow x - 2 \ge 0\;and\,4 - x \ge 0\,\;and\;9 - {x^2} \ge 0$$

$$\Rightarrow x \ge 2\,\,\;{\rm{and}}\,x \le 4\,\,\;{\rm{and}}\,\, - 3 \le x \le 3$$

The intersection of the solution sets of these three inequalities is $$x \in \left[ {2\,,3} \right]\,\,\,\, \Rightarrow \,\,\,D = \left[ {2,\,\,3} \right]$$

(d) $$f(x) = \sqrt {4 - {x^2}} + \sqrt {{x^2} - 4x + 3}$$   $$\Rightarrow \,\,4 - {x^2} \ge 0$$ and $${x^2} - 4x + 3 \ge 0$$

$$\Rightarrow \quad {x^2} \le 4$$and $$(x - 1)(x - 3) \ge 0$$

$$\Rightarrow \quad - 2 \le x \le 2$$ and $$x \le 1$$or $$x \ge 3$$

\begin{align}&\Rightarrow \quad x \in [ - 2,1]\\&\Rightarrow \quad D = [ - 2,1]\end{align}

(e) f(x) = \begin{align}\frac{1}{{\sqrt { - ({x^2} + 3x + 2)} }}\end{align}\,\,\,\,\,\,\, \Rightarrow \,\, - ({x^2} + 3x + 2) > 0\,\, \Rightarrow \,\,{x^2} + 3x + 2 < 0

\begin{align}&\Rightarrow \,\,(x + 1)(x + 2) < 0\\&\Rightarrow \,\, - 2 < x < - 1\\&\Rightarrow \,\,D = ( - 2, - 1)\end{align}

(f) f(x) = \begin{align}\frac{1}{{\sqrt { - ({x^2} + 3x + 2)} }}\end{align} + \sqrt { - ({x^2} - 3x + 2)} \, \Rightarrow ({x^2} + 3x + 2) > 0\,{\rm{and - (}}{x^2} - 3x + 2{\rm{)}} \ge {\rm{0}}

$$\Rightarrow \,\,{x^2} + 3x + 2 < 0$$ and $${x^2} - 3x + 2 \le 0$$

$$\Rightarrow \,\,(x + 1)(x + 2) < 0$$ and $$(x - 1)(x - 2) \le 0$$

$$\Rightarrow \,\, - 2 < x < - 1$$ and $$1 \le x \le 2$$

No value of x can simultaneously satisfy the two constraints above $$\Rightarrow$$ D = $$\phi$$.

(g) $$f(x) = \sqrt {6{x^2} + 5x + 1} \,\, \Rightarrow \,\,6{x^2} + 5x + 1 > 0$$

$$\Rightarrow \,\,(3x + 1)(2x + 1) > 0$$

$$\Rightarrow \,\,x < - \frac{1}{2}$$ or $$x > - \frac{1}{3}$$

$$\Rightarrow \,\,D = \left( { - \infty , - \frac{1}{2}} \right) \cup \left( { - \frac{1}{3},\infty } \right).$$

(h) f(x) = \begin{align}\frac{1}{{\sqrt {100 - {x^2}} }}\end{align} + \sqrt {2 - {x^2}} \,\, \Rightarrow \,\,100 - {x^2} > 0and $$2 - {x^2} \ge 0$$

$$\Rightarrow \,\,{x^2} < 100$$ and $${x^2} \le 2$$

The values of x that would simultaneously satisfy the two inequalities above are

\begin{align}&{x^2} \le 2\,\, \Rightarrow \,\, - \sqrt 2 \le x \le \sqrt 2 \\&\qquad\quad \Rightarrow \,\,D = \left[ { - \sqrt 2 ,\sqrt 2 } \right]\end{align}