Functions - Section V

From LocusWiki

The inverse of a function f(x) is a function g(x) such that if f maps an element ‘a’ to an element ‘b’, g maps ‘b’ to ‘a’.

In more formal terms, $g\left( {f\left( x \right)} \right) = x$

That is, g reverses the action of f on x.

Graph:

An inverse cannot exist for every function. To see why, consider $f:\,\,A \to B$ where A is the domain and B the co-domain. Consider the following maps:

(a) This is an into map (into function). If we take the ‘inverse map’ (from B to A, using the same ‘links’ as in f ), we see that the

element ‘g’ cannot be assigned to any element in A. In other words, for an into function, some values in the co-domain are ‘left out’,

and their ‘inverses’ do not exist.

Graph:

$\Rightarrow$ We cannot defined an inverse for an into function.

(b) This is a many-one map (a many-one function).

If we take the inverse map, the element ‘e’ will be assigned to two elements in A, that is, ‘a’ and ‘b’.

Hence, the inverse map cannot be a function.

Graph:

$\Rightarrow$ We cannot defined an inverse for a many-one function.

From this discussion, we conclude that for a function to be invertible, it should be one-one and onto (also called a bijective function).

Graph:

As is intuitively clear, we can easily define an inverse for the map above.

$\left( {{f^{ - 1}}({\rm{e}}) = {\rm{a}},\,\,{f^{ - 1}}({\rm{f}}) = {\rm{b}},\,\,{f^{ - 1}}({\rm{g}}) = {\rm{c}},\,\,{f^{ - 1}}({\rm{h}}) = {\rm{d}}} \right).$

Lets try to define an inverse for some of the functions that we have seen up till now.

(a) $f\left( x \right) = x$ $\mathbb{R} \to \mathbb{R}$

We see that this function is one-one and onto. The inverse exists.

${f^{ - 1}}(x) = x$ $\mathbb{R} \to \mathbb{R}.$

(b) $f\left( x \right) = {x^2}$ $\mathbb{R} \to \mathbb{R}$

This function is many-one and into. (why?). To define the inverse, we first need to make f one-one and onto.

Graph:

We see that to make the function one-one, we can select the domain as only [0, $\infty$ ) (instead of $\mathbb{R}$ ).

In this domain, the function is one-one, as is clear from the second diagram above.

Also, the range is [0, $\infty$ ). Therefore, redefine this function:

$f\left( x \right) = {x^2}\,\,\,\,\,\,{[0,\infty )}\limits_A \,\,\, \to \,\,\,{[0,\infty )}\limits_B$

This function is now one-one and onto, and hence invertible

${f^{ - 1}}\left( x \right) = \sqrt x \,\,\;{[0,\infty )}\limits_B \,\,\, \to \,\,\,{[0,\infty )}\limits_A$

Additionally, note that if we draw the graph of $\sqrt x$ and ${x^2}$ on the same axis, they are mirror images of each other in the line x = y.

(As we saw in the case of and $\,f(x) = \ln x$ and ${f^{ - 1}}(x) = {e^x})$ .

Graph:

A little thought will show why f and ${f^{ - 1}}$ should be mirror images in the mirror y = x.

In the equation $y = f(x),$ x can be treated as the independent and y the dependent variable.

In the equation $x = {f^{ - 1}}(y)$ , we can reverse the roles. We can treat y-axis as the independent variable axis

Graph:

By convention, the independent variable is taken on the horizontal axis. Therefore we convert this horizontal view (the second diagram above) into a conventional vertical view. How? By taking the reflection in y = x.

Graph:

(c) $f\left( x \right) = \sin x \,\,\,\,\, \mathbb{R} \to \left[ { - 1,\;1} \right]$

Graph:

In other words, we redefine f as $f(x) = \sin x\,\,\,\,\,\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right] \to [ - 1,1]$ as

(we could have taken any other interval in which sin x remains one-one, to define the inverse. Butconventionally, we take the interval closest to the origin; $\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$ in this case, which is centred on the origin). The inverse can now be defined as:

${f^{ - 1}}\left( x \right) = {\sin ^{ - 1}}x\;\;\;\left[ { - 1,1} \right]\; \to \left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$

Graph:

Draw the graphs for ${\cos ^{ - 1}}x$ and ${\tan ^{ - 1}}x$ , after first defining these functions appropriately.

Example - 20

Check whether the following functions are even or odd.

(a) $f(x) = \sin x - \cos x$

(b) $f(x) = \frac{x}{{{e^x} - 1}} + \frac{x}{2} + 1$

(c) $f(x) = \sqrt[3]{{{{(1 - x)}^2}}} + \sqrt[3]{{{{(1 + x)}^2}}}$

Solution

(a) $f( - x) = \sin ( - x) - \cos ( - x) = - \sin x - \cos x$

We see that $f( - x) \ne f\left( { - x} \right)$ and $f( - x)\,\, \ne \,\, - f(x)$

Hence, this function is neither even nor odd.

(b) $f\left( { - x} \right) = \frac{{ - x}}{{{e^{ - x}} - 1}} - \frac{x}{2} + 1\,\,\, = \,\,\,\frac{{x{e^x}}}{{{e^x} - 1}} - \frac{x}{2} + 1$

$= \frac{{x{e^x} + x}}{{2({e^x} - 1)}} + 1\,\,\,\, = \,\,\,\,\frac{{x({e^x} - 1) + 2x}}{{2({e^x} - 1)}} + 1\,\,\,\, = \,\,\,\,\frac{x}{{{e^x} - 1}} + \frac{x}{2} + 1 = f(x)$

Therefore, this function is even.

(c) $f( - x) = f(x)$ , which is obvious just by observation.

This function is even.

Example - 21

If the function f satisfies $f\left( {x + y} \right) + f\left( {x - y} \right) = 2f\left( x \right)\,\,.\,\,f\left( y \right)\,\,<tex>\forall\,\,$ $x,y \,\in \,\mathbb{R},$ and $f\left( 0 \right) \ne 0,$ prove that $f\left( x \right)$ is even

Solution

Substituting x = y = 0,

We get $2f(0) = 2f{(0)^2}\,\,\, \Rightarrow\,\,\, f\left( 0 \right)\left( {f\left( 0 \right) - 1} \right) = 0$

Since $f(0) \ne 0,$ we have $f(0) = 1$

Now substitue $x = 0$

to get $f(y) + f( - y) = 2f\left( 0 \right)f\left( y \right) = 2f\left( y \right)$

$\Rightarrow\,\,\, f\left( y \right) = f\left( { - y} \right)$

$\Rightarrow\,\,\,$ f is even.

Example - 22

If f is non-periodic and g is periodic, what can you say about $f\left( {g(x)} \right)$ and $g\left( {f(x)} \right)$ ?

Solution

Suppose g has a period T,.

Then, $f\left( {g(x + T} \right) = f\left( {g\left( x \right)} \right)$

because $g(x + T) = g\left( x \right)$

Therefore, $f\left( {g(x)} \right)$ is periodic with period T.

$g\left( {f(x)} \right)$ will not be periodic in general. But suppose f(x) is a linear function, that is, f(x) = ax + b.

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

This is periodic with period T / | a |, since g(x) is periodic with period T. (verify)

Example - 23

What are the periods for the following functions?

(a) $f(x) = {\sin ^4}x + {\cos ^4}x$	(b) $f(x) = {\sin ^4}x + {\cos ^4}x$
(c) $f(x) = 2\cos \frac{{x - \pi }}{3}$	(d) $f(x) = {\sin ^2}\pi x$

Solution

The period for $f(x)$ is LCM $\left( {\frac{\pi }{2},\frac{\pi }{3}} \right) = \pi$

(b) ${\sin ^4}x$ and ${\cos ^4}x$ are periodic with period $\pi$ .

Hence f(x) should be periodic with period LCM $(\pi, \pi) = \pi$

But consider $f\left( {x + \frac{\pi }{2}} \right) = {\sin ^4}\left( {x + \frac{\pi }{2}} \right) + {\cos ^4}\left( {x + \frac{\pi }{2}} \right) = {\cos ^4}x + {\sin ^4}x = f(x)$

The fundamental period is therefore not $\pi$ but $\pi/2$ .

(c) This is an example of the case discussed above,{ $g\left( {f(x)} \right),\,\,$ is periodic and f

non-periodic but linear}. The period is $6\pi$ .

$\left\{ {2\cos \left( {\frac{{\left( {x + 6\pi } \right) - \pi }}{3}} \right) = 2\cos \left( {\frac{{x - \pi }}{3} + 2\pi } \right) = 2\cos \left( {\frac{{x - \pi }}{3}} \right)} \right\}$

(d) This is periodic with period 1, since

${\sin ^2}\pi \left( {x + 1} \right) = {\sin ^2}(\pi x + \pi ) = {\sin ^2}\pi x$ .

Example - 24

Prove that if f is increasing, ${f^{ - 1}}$ is increasing and if f is decreasing, ${f^{ - 1}}$ is decreasing.

Solution

If f is increasing, then ${x_1} < {x_2}$ implies that $f({x_1}) < f({x_2})$ or ${y_1} < {y_2}$

Now, just reverse this argument.

For ${y_1} < {y_2},$ we should have ${x_1} < {x_2}$ (from above).

Hence ${y_1} < {y_2} \Rightarrow \,\,{x_1} < {x_2},$ which is equivalent to ${f^{ - 1}}({y_1}) < {f^{ - 1}}({y_2})$

$\therefore$ ${y_1} < {y_2} \Rightarrow {f^{ - 1}}\left( {{y_1}} \right) < {f^{ - 1}}\left( {{y_2}} \right)$

$\Rightarrow {f^{ - 1}}$ is increasing .

The decreasing case can be proved similarly

Note: For students who are not convinced, read the following:

(i) Visualise f in the form of a mapping. If f is increasing,

Graph:

(ii) Consider two statements P and Q, where P implies Q, i.e., $P \Rightarrow Q$ .

Therefore, not $\left( Q \right) \Rightarrow$ not (P)

For example

(Dark clouds are forming in the sky) $\Rightarrow$ (It is about to rain)

(It is not about to rain) $\Rightarrow$ (Dark clouds are not forming in the sky)

Lets apply this to our case. Since f is increasing.

${x_1} < {x_2} \Rightarrow {y_1} < {y_2}$ $\left( {{\rm{where }}y = f\left( x \right),\;x = {f^{ - 1}}\left( y \right)} \right)$

Hence not $\left( {{y_1} < {y_2}} \right) \Rightarrow$ $\left( {{x_1} < {x_2}} \right)$

or ${y_1} > {y_2} \Rightarrow {x_1} > {x_2}$

or ${y_1} > {y_2} \Rightarrow {f^{ - 1}}\left( {{y_1}} \right) > {f^{ - 1}}\left( {{y_2}} \right)$

Hence, ${f^{ - 1}}$ is increasing.

Example - 25

If $f(x) = \sin x + \cos x$ and $g(x) = {x^2} - 1,$ then find the interval in which $g\left( {f(x)} \right)$ is invertible.

Solution

$g\left( {f\left( x \right)} \right) = g(\sin x + \cos x) = {(\sin x + \cos x)^2} - 1$

$= 1 + 2\sin x\cos x - 1 = \sin 2x$

We have seen that sin x is invertible in $\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$ . Therefore, sin 2x will be invertible in $\left[ { - \frac{\pi }{4},\frac{\pi }{4}} \right]$ .

In fact, any function is invertible in all such intervals where it is one-one and onto. sin 2x will be invertible in infinitely many such intervals. But $\left[ { - \frac{\pi }{4},\frac{\pi }{4}} \right]$ can be considered the basic invertible domain for sin 2x since this interval is centered around the origin.

Example - 26

If $f(x + y) = f(x).f(y)\;\forall$ real x, y and $f\left( 0 \right) \ne 0,$ then prove that the function $f(x) = \frac{{f(x)}}{{1 + {{\left( {f(x)} \right)}^2}}}$ is an even function.

Solution

As in Q.14, we first substitute x = y = 0 to get $f\left( 0 \right) = 1 \Rightarrow f\left( x \right) = \frac{1}{{f\left( { - x} \right)}}$

(Note that the approach in such questions is always to try out some sample substitutions that yield significant results).

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

Hence $f\left( x \right)$ is even.

Example - 27

Let f be a real-valued function with domain $\mathbb{R}$ . Now, if for some +ve constant ‘a’, the equation $\[f(x + a) = 1 + {\left\{ {2 - 3f(x) + 3{{\left( {f(x)} \right)}^2} - {{\left( {f(x)} \right)}^3}} \right\}^{{{1/ 3}}$ holds good for all $x \in \mathbb{R}$ , prove that f(x) is a periodic function.

Solution

To prove that f is periodic, one needs to show that f(x + T) = f(x) for some +ve constant T. The equation in the question should give us such a relation,

Although at first sight, this equation seems complicated, a little thought shows that this equation is just an expanded version of the simpler equation:

$f\left( {x + a} \right) = 1 + {\left\{ {1 + {{(1 - f(x))}^3}} \right\}^{{1/{3}}}$

This implies ${\left( {f(x + a) - 1} \right)^3}$ $= 1 + {\left( {1 - f(x)} \right)^3}$

$\Rightarrow$ ${\left( {f(x) - 1} \right)^3} + {\left( {f(x + a) - 1} \right)^3} = 1$ .... (i)

To proceed, we can substitute x + a for x, since the equation is valid for all x.

$\Rightarrow$ ${\left( {f(x + a) - 1} \right)^3} + {\left( {f(x + 2a) - 1} \right)^3} = 1$ ... (ii)

Substitution of (i) in (ii) yields $f\left( {x + 2a} \right) = f\left( x \right)$

Therefore, f is periodic with period ‘2a’

Functions - Section V

From LocusWiki

Example - 20

Solution

Example - 21

Solution

Example - 22

Solution

Example - 23

Solution

Example - 24

Solution

Example - 25

Solution

Example - 26

Solution

Example - 27

Solution

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