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Examples on Domains and Ranges of Functions Set 6

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Example- 20

Find the domains of the following functions

(i) f\left( x \right) = \begin{align}\sqrt {\frac{{{x^2} + 1}}{{{x^2} + x + 1}}} - \frac{1}{{{{\left[ x \right]}^2} + 3\left[ x \right] + 2}} + {3^{\sqrt {5 - {x^2}} }}\end{align}               (ii) f\left( x \right) = \begin{align}\frac{1}{{{{\left\{ x \right\}}^2} - 1/4}} + \frac{1}{{\left\{ x \right\}}}\end{align}

Solution:

(i) f\left( x \right) =\begin{align} \sqrt {\frac{{{x^2} + 1}}{{{x^2} + x + 1}}} - \frac{1}{{{{\left[ x \right]}^2} + 3\left[ x \right] + 2}} + {3^{\sqrt {5 - {x^2}} }}\end{align}

\begin{align}\Rightarrow \frac{{{x^2} + 1}}{{{x^2} + x + 1}} \ge 0,\,\,\,\,\,\,\,{\left[ x \right]^2} + 3\left[ x \right] + 2 \ne 0,\,\,\,\,\,\,\,\,\,\,\,\,5 - {x^2} \ge 0\end{align}

Observe that the first constraint is always satisfied since both the numerator and the denominator are positive for all values of x. The other two constraints are:

\begin{align}&\left( {\left[ x \right] + 1} \right)\left( {\left[ x \right] + 2} \right) \ne 0,\,\,\,\,\,{x^2} \le 5\\&\Rightarrow \left[ x \right] \ne - 1, - 2\,\,\,\, - \sqrt 5 \le x \le \sqrt 5 \\&\Rightarrow x \in \left[ { - \sqrt 5 , - 2} \right) \cup \left[ {0,\sqrt 5 } \right)\,\,\,\,\,\, \Rightarrow D = \left[ { - \sqrt 5 , - 2} \right) \cup \left[ {0,\sqrt 5 } \right)\\\end{align}

(ii) f\left( x \right) =\begin{align} \frac{1}{{{{\left\{ x \right\}}^2} - \frac{1}{4}}} + \frac{1}{{\left\{ x \right\}}} \Rightarrow \left\{ x \right\} \ne 0,\frac{1}{2},\frac{{ - 1}}{2}\end{align}

$$\begin{gathered}\Rightarrow x \ne n,\,\,\,n + \frac{1}{2};\,\,\,\,\,\,\,\,\,\,n \in \mathbb{Z} \hfill \\\Rightarrow D = \mathbb{R}\backslash \left\{ {n,n + \frac{1}{2}} \right\};\,\,\,\,\,\,\,\,n \in \mathbb{Z} \hfill \\\end{gathered}$$

Example- 21

Find the range of $$f(x)$$ = \begin{align}\frac{{{x^2} - 3x + 2}}{{{x^2} + x - 6}}\,\end{align}

Solution: First, we find out the domain. We require

\begin{align}&{x^2} + x - 6 \ne 0\\&\Rightarrow x \ne - 3,2\\&D = \mathbb{R} - \left\{ { - 3,2} \right\} \end{align}

Now \begin{align}f(x) =\frac{{(x - 1)(x - 2)}}{{(x + 3)(x - 2)}} = \frac{{x - 1}}{{x + 3}}\end{align}

(This cancellation is allowed if x $$\ne$$ 2, which we have ensured in the domain)

$$x \ne 2\;implies\;f(x) \ne 1/5$$

Now we write x in terms of f(x) (we will soon see why)

$$x f\left( x \right){\rm{ }} + {\rm{ }}3f\left( x \right){\rm{ }} = x-{\rm{ }}1$$

$$\Rightarrow$$ x = \begin{align}\frac{{3f(x) + 1}}{{1 - f(x)}}\end{align}

We require that x be real, for which the right hand side should be defined. This requirement will place a restriction on the values that f(x) can take. For all the allowed values of f(x), there will exist a valid value of x (a pre-image’ l, or an ‘inverse’). For unallowed values of f(x), there will exist no pre-image or no inverse. Hence f(x) will not take on such values, or in other words, such values will not lie in the range of f. In the expression above, we see that f(x) $$\ne$$1.

Hence, the range is R = $$\mathbb{R} - \left\{ {\frac{1}{5},\,\,1} \right\}$$

Note: The method above is a standard method to evaluate the range, if x can be written in terms of f(x).

$x = (g\left( {f(x)} \right)$

The domain for g (the values for which g is defined) is the set of values that f can take, and hence, is the range of f.

Example- 22

Find the range of (a) f\left( x \right) = \begin{align}\frac{{{x^2} + x - 1}}{{{x^2} - x + 2}}\end{align} (b) f\left( x \right) = \begin{align} \frac{{{x^2} + x + 1}}{{{x^2} + x + 2}}\,\end{align}

Solution:

(a) For the domain,$${x^2} - x + 2 \ne 0$$

$${x^2} - x + 2$$can be rearranged as $${\left( {x - \frac{1}{2}} \right)^2} + 7/4$$, which is always positive

Hence, for no value of x is the denominator 0.

$\Rightarrow D =\mathbb{R}$

To find the range, we find x in terms of f(x), as in the previous question, and see what values of f(x) will make x real. Those values will form our range.

(We use y instead of f(x) for convenience i.e. y = f(x))

$${x^2}y - xy + 2y$$ = $${x^2} + x - 1$$

$$\Rightarrow$$ $$(1 - y){x^2} + (1 + y)x - (1 + 2y)$$= 0 (*)

By the quadratic formula.

\begin{align}&x = \frac{{ - (1 + y) \pm \sqrt {{{(1 + y)}^2} + 4(1 - y)(1 + 2y)} }}{{2(1 - y)}}\\&x = \frac{{ - (1 + y) \pm \sqrt { - 7{y^2} + 6y + 5} }}{{2(1 - y)}}\end{align}

The right side is defined when y$$\ne$$1 and > 0 $$- 7{y^2} + 6y + 5$$ $$\Rightarrow$$ $$7{y^2} - 6y - 5 \le 0$$

$\Rightarrow {\rm{ }}\frac{{3 - 2\sqrt {11} }}{7} \le y \le {\rm{ }}\frac{{3 + 2\sqrt {11} }}{7}$

Therefore, the values that y can take are

$\frac{{3 - 2\sqrt {11} }}{7} \le y \le {\rm{ }}\frac{{3 + 2\sqrt {11} }}{7}\quad and \quad{\rm{ }}y{\rm{}} \ne 1$

These values should form our range. But we have to be a little careful here. y $$\ne$$ 1 arises because the quadratic formula gives (1 – y) in the denominator. Suppose that in the (*) equation itself, we put y = 1, reducing the equation to a linear one:

$$2x - 3 = 0{\rm{ }} \Rightarrow x = 3/2$$

This means that y = 1 gives a defined value of x, or in other words, y =1 has a pre-image, x = 3/2. $$\left( {{\rm{check this : }}f\left( {\frac{3}{2}} \right) = 1} \right)$$

Hence y =1 should also be in the range.

$R = {\rm{ }}\left[ {\frac{{3 - 2\sqrt {11} }}{7},\frac{{3 + 2\sqrt {11} }}{7}} \right]$

(b)  f\left( x \right) =\begin{align} \frac{{{x^2} + x + 1}}{{{x^2} + x + 2}}\end{align}

To evaluate the range of such (rational) expressions whose range is not obtainable from simple rearrangement (like in a quadratic expression), we put the expression (in x) equal to some variable y, write x as a function of y (x = g(y)) and find the domain of g; this domain consists of values that y can take or our required range (An example of this sort is described in the unit on functions).

$\Rightarrow y = \frac{{{x^2} + x + 1}}{{{x^2} + x + 2}} \Rightarrow \left( {1 - y} \right){x^2} + \left( {1 - y} \right)x + \left( {1 - 2y} \right) = 0$

(Note that y can never equal 1)

Solving x in terms of y, we get

$x = \frac{{ - \left( {1 - y} \right) \pm \sqrt {{{\left( {1 - y} \right)}^2} - 4\left( {1 - y} \right)\left( {1 - 2y} \right)} }}{{2\left( {1 - y} \right)}}$

For x to be real,

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

\begin{align}&\Rightarrow {y^2} - 2y + 1 - 4\left( {2{y^2} - 3y + 1} \right) \ge 0\\&\Rightarrow 7{y^2} - 10y + 3 \le 0\\&\Rightarrow \left( {7y - 3} \right)\left( {y - 1} \right) \le 0.\\&\Rightarrow \frac{3}{7} \le y \le 1\end{align}

Because $$y \ne 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow R = \left[ {\frac{3}{7},1} \right)$$

Observe that the range could; in this particular example, have been evaluated in a simple manner as follows (by rearrangement of f(x))

$f\left( x \right) = 1 - \frac{1}{{{x^2} + x + 2}} = 1 - \frac{1}{{{{\left( {x + \frac{1}{2}} \right)}^2} + \frac{7}{4}}}$

Now 0 <\begin{align} \frac{1}{{{{\left( {x + \frac{1}{2}} \right)}^2} + 7/4}} \le \frac{4}{7}\,\,\,\,\, \Rightarrow \frac{3}{7} \le f\left( x \right) < 1\,\,\, \Rightarrow R = \left[ {\frac{3}{7},1} \right)\end{align}

We followed the first approach to describe the general outline used to solve such cases.

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