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Question: If we have a function \[f\left( x \right)=\dfrac{5x+6}{7x+9}\] then \[{{f}^{-1}}\left( x \right)=\] ...

If we have a function f(x)=5x+67x+9f\left( x \right)=\dfrac{5x+6}{7x+9} then f−1(x)={{f}^{-1}}\left( x \right)=
(A). y+67y+9\dfrac{y+6}{7y+9}
(B). 7y+95y+6\dfrac{7y+9}{5y+6}
(C). 9y−6−7y+9\dfrac{9y-6}{-7y+9}
(D). 9y−6−7y+5\dfrac{9y-6}{-7y+5}

Explanation

Solution

Hint: First equate the given function to y. Now you have a relation between y and x. y is on the left-hand side and x on the right-hand side. Now apply cross multiplication or in normal terms multiply the denominator on both sides. Now you have an equation with x terms on both sides. Now find the value of x term on the right-hand side. Subtract this term on both sides of the equation. Now find the value of y term on the left-hand side. Subtract this term on both sides of the equation. Now you have all x terms on the left-hand side. Just take the x common on the left-hand side. Now divide with the coefficient of this x which is a variable equation in terms of y. Now you get the value of x which is nothing but inverse.

Complete step-by-step solution -
Assume the function as y, then you can write as below:
f (x) = y
Now apply f−1{{f}^{-1}} on both sides of above equation, we get:
f−1(f(x))=f−1(y){{f}^{-1}}\left( f\left( x \right) \right)={{f}^{-1}}\left( y \right)
By basic knowledge, we can say that below equation as:
x=f−1(y)x={{f}^{-1}}\left( y \right) ------- (1)
Given equation in the question, after assuming it as y is:
y=5x+67x+9y=\dfrac{5x+6}{7x+9}
By multiplying with (7x+9)\left( 7x+9 \right) on both sides, we get it as:
⇒y(7x+9)=5x+6\Rightarrow y\left( 7x+9 \right)=5x+6
By simplifying the term on left hand side, we get it as:
7xy + 9y = 5x + 6
By subtracting the term 5x on both sides, we get it as:
7xy + 9y – 5x = 6
By subtracting the term 9y on both sides, we get it as:
7xy – 5x = 6 – 9y
By taking x common from the term on left hand side, we get:
x (7y - 5) = 6 – 9y
By dividing with (7y - 5) on both sides, we get it as:
⇒x=6−9y7y−5\Rightarrow x=\dfrac{6-9y}{7y-5}
By substituting equation (1) here, we get: f−1(y)=6−9y7y−5{{f}^{-1}}\left( y \right)=\dfrac{6-9y}{7y-5}.
By above, we can get: f−1(x)=6−9x7x−5=9x−6−7x+5{{f}^{-1}}\left( x \right)=\dfrac{6-9x}{7x-5}=\dfrac{9x-6}{-7x+5}.
Therefore option (d) is correct.

Note: Generally students confuse between x, y. y is a function in terms of x. Remember that the value of x in terms of the function itself is nothing but its inverse. Be careful while removing brackets, students generally forget to write 9y which will affect the final result. The step of multiplying denominators is also called cross-multiplication. Do it carefully. Don’t make mistakes in it.