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We must also understand that this new function, “ ${f}^{-1}$ ”, gives the perspective of relation from co-domain to domain of the given function “f”. However, new function “ ${f}^{-1}$ ” is read from its new domain to its new co-domain. After all this is how a function is read. This simply means that domain and co-domain of the function “f” is exchanged for “ ${f}^{-1}$ ”.
Further, inverse function is inverse of a given function. Again by definition, every element of domain set of the given function “f” is also related to exactly one element of in its co-domain. Thus, there is bidirectional requirement that elements of one set are related to exactly one element of other set. Clearly, this requirement needs to be fulfilled, before we can define inverse function.
In other words, we can define inverse function, " ${f}^{-1}$ ", only if the given function is an injection and surjection function (map or relation) at the same time. Hence, iff function, “f” is a bijection, then inverse function is defined as :
$${f}^{-1}:A\to B\phantom{\rule{1em}{0ex}}\text{by}\phantom{\rule{1em}{0ex}}{f}^{-1}\left(x\right)\phantom{\rule{1em}{0ex}}\text{for all}\phantom{\rule{1em}{0ex}}x\in A$$
We should again emphasize here that sets “A” and “B” are the domain and co-domain respectively of the inverse function. These sets have exchanged their place with respect to function “f”. This aspect can be easily understood with an illustration. Let a function “f” , which is a bijection, be defined as :
Let A = {1,2,3,4} and B={3,6,9,12}
$$f:A\to B\phantom{\rule{1em}{0ex}}\text{by}\phantom{\rule{1em}{0ex}}f\left(x\right)=3x\phantom{\rule{1em}{0ex}}\text{for all}\phantom{\rule{1em}{0ex}}x\in A$$
Then, the function set in the roaster form is :
$$\Rightarrow f=\{\left(\mathrm{1,3}\right),\left(\mathrm{2,6}\right),\left(\mathrm{3,9}\right),\left(\mathrm{4,12}\right)\}$$
This function is clearly a bijection as only distinct elements of two sets are paired. Its domain and co-domains are :
$$\Rightarrow \text{Domain of \u201cf\u201d}=\left\{\mathrm{1,2,3,4}\right\}$$
$$\Rightarrow \text{Co-domain of \u201cf\u201d}=\left\{\mathrm{3,6,9,12}\right\}$$
Now, the inverse function is given by :
$${f}^{-1}:A\to B\phantom{\rule{1em}{0ex}}by\phantom{\rule{1em}{0ex}}f\left(x\right)=\frac{x}{3}\phantom{\rule{1em}{0ex}}\text{for all}\phantom{\rule{1em}{0ex}}x\in A$$
$$\phantom{\rule{1em}{0ex}}\text{where}\phantom{\rule{1em}{0ex}}A=\left\{\mathrm{3,6,9,12}\right\}$$
In the roaster form, the inverse function is :
$$\Rightarrow {f}^{-1}=\{\left(\mathrm{3,1}\right),\left(\mathrm{6,2}\right),\left(\mathrm{9,3}\right),\left(\mathrm{12,4}\right)\}$$
Note that we can find inverse relation by merely exchanging positions of elements in the ordered pairs. The domain and co-domain of new function “ ${f}^{-1}$ ” are :
$$\Rightarrow \text{Domain of}\phantom{\rule{1em}{0ex}}{f}^{-1}=\left\{\mathrm{3,6,9,12}\right\}$$
$$\Rightarrow \text{Co\u2212domain of}\phantom{\rule{1em}{0ex}}{f}^{-1}=\left\{\mathrm{1,2,3,4}\right\}$$
Thus, we see that the domain of inverse function “ ${f}^{-1}$ ” is co-domain of the function “f” and co-domain of inverse function “ ${f}^{-1}$ “ is domain of the function “f”.
Problem 1: A function is given as :
$$f:R\to R\phantom{\rule{1em}{0ex}}\text{by}\phantom{\rule{1em}{0ex}}f\left(x\right)=2x+5\phantom{\rule{1em}{0ex}}\text{for all}\phantom{\rule{1em}{0ex}}x\in R$$
Construct the inverse rule. Determine f(x) for first 5 natural numbers. Check validity of inverse rule with the values of images so obtained. Find inverse function, if it exists.
Solution : Following the illustration given earlier, we derive inverse rule as :
$$y=2x+5$$
$$\Rightarrow x=\frac{y-5}{2}$$
Changing notations,
$$\Rightarrow {f}^{-1}\left(x\right)=\frac{x-5}{2}$$
The images i.e. corresponding f(x), for first five natural numbers are :
$$f\left(1\right)=2x+5=7;\phantom{\rule{1em}{0ex}}f\left(2\right)=9;\phantom{\rule{1em}{0ex}}f\left(3\right)=11;\phantom{\rule{1em}{0ex}}f\left(4\right)=13\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}f\left(5\right)=15$$
Now, the corresponding pre-images, using inverse rule for two values of images are :
$$\Rightarrow {f}^{-1}\left(7\right)=\frac{7-5}{2}=1$$
$$\Rightarrow {f}^{-1}\left(11\right)=\frac{11-5}{2}=3$$
Thus, we see that the inverse rule correctly determines the pre-images as intended. Now, in order to find inverse function, we need to determine that the given function is an injection and surjection. For injection, let us assume that “ ${x}_{1}$ ” and “ ${x}_{2}$ ” be two different elements such that :
$$f\left({x}_{1}\right)=f\left({x}_{2}\right)$$
$$\Rightarrow 2{x}_{1}+5=2{x}_{2}+5$$
$${x}_{1}={x}_{2}$$
This means that given function is an injection. Now, to prove surjection, we solve the rule for “x” as :
$$\Rightarrow x=\frac{y-5}{2}$$
We see that this equation is valid for all values of “R” i.e. all values in the co-domain of the given function. This means that every element of the co-domain is related. Hence, given function is surjection. The inverse function, therefore, is given as :
$${f}^{-1}:R\to R\phantom{\rule{1em}{0ex}}by\phantom{\rule{1em}{0ex}}f\left(x\right)=\frac{\left(x-5\right)}{2}\phantom{\rule{1em}{0ex}}\text{for all}\phantom{\rule{1em}{0ex}}x\in R$$
There are few characteristics of inverse function that results from the fact that it is inverse of a bijection. We can check the validity of these properties in terms of the example given earlier. Let us define a bijection function as defined earlier :
$$\mathrm{Let}\phantom{\rule{1em}{0ex}}A=\left\{\mathrm{1,2,3,4}\right\}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}B=\left\{\mathrm{3,6,9,12}\right\}$$
$$f:A\to B\phantom{\rule{1em}{0ex}}\text{by}\phantom{\rule{1em}{0ex}}f\left(x\right)=3x\phantom{\rule{1em}{0ex}}\text{for all}\phantom{\rule{1em}{0ex}}x\in A$$
This means that there is only one inverse function. For the given function the inverse function is :
$${f}^{-1}:A\to B\phantom{\rule{1em}{0ex}}by\phantom{\rule{1em}{0ex}}f\left(x\right)=\frac{x}{3}\phantom{\rule{1em}{0ex}}forall\phantom{\rule{1em}{0ex}}x\in A$$
$$\phantom{\rule{1em}{0ex}}\text{where}\phantom{\rule{1em}{0ex}}A=\left\{\mathrm{3,6,9,12}\right\}$$
In the roaster form, the inverse function is :
$$\Rightarrow {f}^{-1}=\{\left(\mathrm{3,1}\right),\left(\mathrm{6,2}\right),\left(\mathrm{9,3}\right),\left(\mathrm{12,4}\right)\}$$
This inverse function is unique to a given bijection.
We see that inverse comprises of ordered pairs such that elements of domain and co-domain are distinctly related to each other.
$$\Rightarrow {f}^{-1}=\{\left(\mathrm{3,1}\right),\left(\mathrm{6,2}\right),\left(\mathrm{9,3}\right),\left(\mathrm{12,4}\right)\}$$
This mean that the inverse function is bijection.
If a function is bijection, then the inverse of function exists. On the other hand, a function is bijection, if it is both one-one and onto function. We know that one-one function is strictly monotonic in its domain. Hence, an onto function is invertible, if its graph is strictly monotonic i.e. either increasing or decreasing.
In order to investigate the nature of the inverse graph, let us consider a plot of an invertible function, “f(x)”. Let (a,b) be a point on the plot. Then, by definition of an inverse function, the point (b,a) is a point on the plot of inverse function, if plotted on the same coordinate system.
$$y=f\left(x\right)$$
$$y={f}^{-1}\left(x\right)$$
By geometry, the line joining points (a,b) and (b,a) is bisected at right angles by the line y = x. It means that two points under consideration are object and image for the mirror defined by y=x. This relationship also restrains that two plots can intersect only at line y = x.
Problem 2 : Two functions, inverse of each other, are given as :
$$f\left(x\right)={x}^{2}-x+1$$
$${f}^{-1}\left(x\right)=\frac{1}{2}+\sqrt{\left(x-\frac{3}{4}\right)}$$
Find the solution of the equation :
$${x}^{2}-x+1=\frac{1}{2}+\sqrt{\left(x-\frac{3}{4}\right)}$$
Solution :
Statement of the problem : The given functions are inverse to each other, which can intersect only at line defined by y = x. Clearly, the intersection point is the solution of the equation.
$$y=f\left(x\right)=x$$
$$\Rightarrow {x}^{2}-x+1=x\phantom{\rule{1em}{0ex}}\Rightarrow {x}^{2}-2x+1=0$$
$$\Rightarrow {\left(x-1\right)}^{2}=0$$
$$\Rightarrow x=1$$
This is the answer. It is interesting to know that we can also proceed to find the solution by working on the inverse function. This should also give the same result as given functions are inverse to each other.
$$y={f}^{-1}\left(x\right)=x$$
$$\Rightarrow \frac{1}{2}+\sqrt{\left(x-\frac{3}{4}\right)}=x$$
$$\Rightarrow \sqrt{\left(x-\frac{3}{4}\right)}=x-\frac{1}{2}$$
Squaring both sides,
$$\Rightarrow \left(x-\frac{3}{4}\right)={\left(x-\frac{1}{2}\right)}^{2}={x}^{2}+\frac{1}{4}-x$$
$$\Rightarrow {x}^{2}+\frac{1}{4}-x-x+\frac{3}{4}=0\phantom{\rule{1em}{0ex}}\Rightarrow {x}^{2}-2x+1=0$$
$$\Rightarrow x=1$$
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