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Question: If we have a function as \(f\left( x \right) = \dfrac{{4x + 3}}{{6x - 4}},x \ne \dfrac{2}{3}\) Show ...

If we have a function as f(x)=4x+36x4,x23f\left( x \right) = \dfrac{{4x + 3}}{{6x - 4}},x \ne \dfrac{2}{3} Show that fof(x)=xfof\left( x \right) = x for all x23x \ne \dfrac{2}{3}. What is the inverse of ff?

Explanation

Solution

Before attempting this question one should have prior knowledge about the concept of functions and also remember that fof(x)=xfof\left( x \right) = x means f(f(x))=xf\left( {f\left( x \right)} \right) = x so to proof this use f(f(x))=f(4x+36x4)f\left( {f\left( x \right)} \right) = f\left( {\dfrac{{4x + 3}}{{6x - 4}}} \right), use this information to approach the solution.

Complete step-by-step solution:
According to the given information we have function f(x)=4x+36x4,x23f\left( x \right) = \dfrac{{4x + 3}}{{6x - 4}},x \ne \dfrac{2}{3}
First of all, we have to show fof(x)=xfof\left( x \right) = x which also means f(f(x))=xf\left( {f\left( x \right)} \right) = x
So, f(f(x))=f(4x+36x4)f\left( {f\left( x \right)} \right) = f\left( {\dfrac{{4x + 3}}{{6x - 4}}} \right)
Now, substituting the value of f(x)f\left( x \right)i.e. 4x+36x4\dfrac{{4x + 3}}{{6x - 4}}
f(f(x))=f(4x+36x4)=4(4x+36x4)+36(4x+36x4)4f\left( {f\left( x \right)} \right) = f\left( {\dfrac{{4x + 3}}{{6x - 4}}} \right) = \dfrac{{4\left( {\dfrac{{4x + 3}}{{6x - 4}}} \right) + 3}}{{6\left( {\dfrac{{4x + 3}}{{6x - 4}}} \right) - 4}}
\Rightarrow f(f(x))=16x+12+18x126x424x+1824x+166x4f\left( {f\left( x \right)} \right) = \dfrac{{\dfrac{{16x + 12 + 18x - 12}}{{6x - 4}}}}{{\dfrac{{24x + 18 - 24x + 16}}{{6x - 4}}}}
\Rightarrow f(f(x))=16x+12+18x1224x+1824x+16f\left( {f\left( x \right)} \right) = \dfrac{{16x + 12 + 18x - 12}}{{24x + 18 - 24x + 16}}
\Rightarrow f(f(x))=34x34=xf\left( {f\left( x \right)} \right) = \dfrac{{34x}}{{34}} = x
Hence proved, f(f(x))=xf\left( {f\left( x \right)} \right) = x
Let ggbe the inverse of the given function
So, we know that if g is inverse of function f(x)f\left( x \right)
Then gof(x)=xgof\left( x \right) = x and fog(x)=xfog\left( x \right) = x
Now comparing the above statement with fof(x)=xfof\left( x \right) = x
So, as we can say that after comparing gof(x)=xgof\left( x \right) = x with fof(x)=xfof\left( x \right) = x
Here g is ff
And on comparing fog(x)=xfog\left( x \right) = x with fof(x)=xfof\left( x \right) = x again here f(x)f\left( x \right) is g(x)g\left( x \right)
Therefore, we can say that the given function is inverse of itself
Which means f(x)=f1(x)f\left( x \right) = {f^{ - 1}}\left( x \right)
You can easily see inverse of ff is equal to f(x)f\left( x \right)
Hence, f1(x)=4x+36x4{f^{ - 1}}\left( x \right) = \dfrac{{4x + 3}}{{6x - 4}}

Note: In the above solution we came across the term “function” which can be explained as a relation between the provided inputs and the outputs of the given inputs such that each input is directly related to the one output. The representation of a function is given by supposing if there is a function “f” that belongs from X to Y then the function is represented by f:XYf:X \to Y examples of function are one-one functions, onto functions, bijective functions, trigonometric function, binary function, etc.