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

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

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

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