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Question: What is inverse function? And can you please explain about the inverse function with the solved exam...

What is inverse function? And can you please explain about the inverse function with the solved example.

Explanation

Solution

If we are given that y=f(x)y = f(x) then the inverse function would be x=f1(y)x = {f^{ - 1}}(y) so inverse function is the function such that if we take the positive sign then the negative sign will be our inverse of it and so on.

Complete step-by-step answer:
An inverse function is also known as the anti-function which is defined as the function which can reverse into another function that means if the function accepts the certain value or perform the particular operation on the values and generates an output then the inverse function agrees with the resultant, operates and reaches back to the original function in simple mathematics. Let us suppose any function ff takes xx to yy. That means that for any domain xx the range will be yy and so if ff is the function, thenf1{f^{ - 1}} is known as its inverse. If we have the function y=f(x)y = f(x) then the inverse function would be x=f1(y)x = {f^{ - 1}}(y)
Let us understand it by an example. Let us suppose that the question is to find the inverse of the function f(x)=2x+3f(x) = 2x + 3
Here function ff is defined as f(x)=2x+3f(x) = 2x + 3
Now we need to find the inverse of this function f(x)=2x+3f(x) = 2x + 3
Therefore let us assume that y=f(x)y = f(x)
y=2x+3y = 2x + 3
So then we can calculate xx
\Rightarrow y3=2xy - 3 = 2x
\Rightarrow x=y32x = \dfrac{{y - 3}}{2}
Here we get the value of xx in terms of yy
As we know that y=f(x)y = f(x)
Then x=f1(y)x = {f^{ - 1}}(y)
Therefore f1(y)=y32{f^{ - 1}}(y) = \dfrac{{y - 3}}{2}
Or we can say that f1(x)=x32{f^{ - 1}}(x) = \dfrac{{x - 3}}{2}
Therefore for f(x)=2x+3f(x) = 2x + 3 its inverse would be f1(x)=x32{f^{ - 1}}(x) = \dfrac{{x - 3}}{2}.

Note: If we are given that g(x)g(x) is the inverse of f(x)f(x) then we can write that
g(x)=f1(x)g(x) = {f^{ - 1}}(x) or we can also write it as f(g(x))=xf(g(x)) = x and for f(g(x))f(g(x)) we take it as fog(x)fog(x)
So we can write it as fog(x)=xfog(x) = x