Examples on Exponential and Logarithmic Functions Set 2

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Example- 28

Find the domains for the function.

\[f(x) = \sin \left( {\ln \left( {\frac{{\sqrt {4 - {x^2}} }}{{1 - x}}} \right)} \right)\]

Solution: This is an example of a composition of functions.

\[f\left( {g\left( {h..........(x)} \right)} \right)\]

The previous question was a composition of four ‘log’ functions

Suppose we have a composite function \(z(x) = f\left( {g\left( {h(x)} \right)} \right)\). Its is helpful to visualise z(x) in the following manner:

To determine the domain of \(z(x),\) it is intutive that we start from the rightmost side (or the outermost function, \(f\) ) and analyses what values \(g(h(x))\) (the domain for \(f\) ) can take; and then accordingly what values \(h(x)\ {the domain for \(g\)} can take, and finally, what values \(x\) can take.

Visualizing the example of the previous question in this manner, we get

We require \({y_3} > {\rm{ }}0\) (the domain for the outermost function, log2)

This gives \({y_2} > {\rm{ }}1{\rm{ }}\) (the domain for log3) and so on.

Applying this concept to our current example, we see that the domain for the ‘sin’ function is \(\mathbb{R}\)

Therefore, \(\ln \left( {\frac{{\sqrt {4 - {x^2}} }}{{1 - x}}} \right) \in \,\mathbb{R}\)

Now, the domain for the ‘ln’ function is \({\mathbb{R}^ + }\) (positive real numbers)

Hence we require \(\begin{align}\frac{{\sqrt {4 - {x^2}} }}{{1 - x}} > 0\end{align}\)

The numerator is defined for \(x \in [ - 2,\;2],\) the denominator for \(x \ne 1.\) Also, the numerator is always positive (except when \(x=\pm 2,\) when it becomes 0) (square root function yields non-negative output by definition)

The denominator is positive when \(x < 1.\)

Therefore, \(x\) should satisfy the following conditions:

\[x{\mkern 1mu}  \in {\mkern 1mu} [ - 2,2]{\rm{  }}\quad and \quad {\rm{ x }} \ne 1{\rm{  }}\quad and \quad {\rm{  x }} \ne  \pm 2{\rm{  }}\quad and  \quad {\rm{  }}x < 1\]

\[ \Rightarrow {\mkern 1mu} D = ( - 2,1)\]

Example- 29

Find the domain and range of

(a) \(f(x) = {e^{\sin x}}\)                                    (b) \(f(x)x = \sin ({e^x})\)

(c) \(f(x) = \ln \,(\sin ({e^x}))\)                          (d) \(f(x) = \ln (\sin (\ln \,x))\)

Solution: (a) As \(x\) varies over \(\mathbb{R}\), \(sin\,x\) varies in [–1, 1], so \({e^{\sin x}}\) will vary from a minimum value of \(\begin{align}{e^{ - 1}} = \frac{1}{e}\end{align}\) to a maximum value of \({e^1} = e\). Let us depict this graphically.

Therefore,

\(\begin{gathered} D = \mathbb{R} \hfill \\ R = \left[ {\;\frac{1}{e},\;e\;} \right] \hfill \\ \end{gathered} \)

(b) As \(x\) varies over \(\mathbb{R},\;{e^x}\) varies in \((0,\;\infty )\). The range of \(\sin ({e^x})\) will be the range of the sin function: [–1, 1]. Convince yourself about this:

Therefore,

\(\begin{gathered}D = \mathbb{R} \hfill \\R = [ - 1,\;1] \hfill \\\end{gathered} \)

(c) In this case, we’ll have a constraint on the domain:

\(f(x) = l\,n\,\left( {\sin ({e^x})} \right)\)

\({\rm{}} \Rightarrow \quad \sin \left( {{e^x}} \right){\rm{}}must\;be\;positive\;for\;ln\;to\;be\;defined\)

\( \Rightarrow \qquad \sin \left( {{e^x}} \right) > 0\)

What values can \({e^x}\) take then for \(\sin \left( {{e^x}} \right)\) to be positive? Well, \({e^x}\) can lie in \((0,\;\pi ),\;(2\pi ,\;3\pi ),\;(4\pi ,\;5\pi ).......\)and generally in \(\left( {2n\pi ,\;(2n + 1)\pi } \right),\;n \ge 0\). Look at the graph of \(sin\,x\) and convince yourself that \(sin\,x\) is positive in these intervals. {Intervals on the negative side have not been taken since \({e^x}\) cannot be negative}. So,

If \({e^x} \in (0,\;\pi ) \qquad \Rightarrow \qquad x \in ( - \infty ,\;l\,n\;\pi )\)

If \({e^x} \in (2\pi ,\;3\pi ) \qquad \Rightarrow \qquad x \in (l\,n\;2\pi ,\;l\,n\;3\pi )\)

Generalizing, the set of allowed values for \(x,\) i.e., the domain, is all intervals of the form

\(\left( {l\,n\;2n\pi ,\;l\,n(2n + 1)\pi } \right),\;\;\;n \ge 0,\;\;n \in \mathbb{Z}\)

What is the range? For the given domain \(\sin ({e^x})\), will take positive values, but \(\sin ({e^x})\) cannot rise above 1. Thus,

\(\begin{align}&\qquad\qquad\quad\sin ({e^x}) \in (0,\;1]\\&\Rightarrow \qquad \ln \,(\sin ({e^x})) \in (-\infty ,\;0]\end{align}\)

The range is \(( -\infty ,\;0]\)

(d) We are only providing the answers. The justification is left to the reader as an exercise:

D = All intervals of the form \(\left( {{e^{2n\pi }},\;{e^{(2n + 1)\pi }}} \right),\;n \in \mathbb{Z}\)

\(R = \left( { - \infty ,\;0} \right]\)

TRY YOURSELF - III

Q.1 Find the domains and range of the following functions:

(a) \(f(x) = {e^{\{ x\} }}\)  (b) \(f(x) = ln\,\left( {\sqrt x } \right)\)
(c) \(f(x) = {e^{[x]}}\)     (d) \(f(x) = {e^{\tan x}}\)
(e) \(f(x) = ln\left( {{x^2} + 1} \right)\)  (f) \(f(x) = ln\{ x\} \)