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Question: If \(f\left( x \right)=3x-5\), then find \({{f}^{-1}}\left( x \right)\). A. \(\dfrac{1}{3x-5}\) ...

If f(x)=3x−5f\left( x \right)=3x-5, then find f−1(x){{f}^{-1}}\left( x \right).
A. 13x−5\dfrac{1}{3x-5}
B. x+53\dfrac{x+5}{3}
C. does not exist because f is not one-one.
D. does not exist because f is not onto.

Explanation

Solution

We first need to describe how and why f(x)=3x−5f\left( x \right)=3x-5 is a function. We find the characteristics of the function f(x)=3x−5f\left( x \right)=3x-5. Then we find the inverse of the function by expressing x as a relation of y. We check if that relation is a function or not.

Complete step by step answer:
We have been given a function in the form of y=f(x)=3x−5y=f\left( x \right)=3x-5.
A function exits only when the function is not one to many. It has to get its fixed image for a particular preimage.
In our given function f(x)=3x−5f\left( x \right)=3x-5, the function is one-one function. This means if a and b belong to the domain and a=ba=b then obviously f(a)=f(b)f\left( a \right)=f\left( b \right).
We need to find the function f−1(x){{f}^{-1}}\left( x \right).
We try to express x with respect to y from the relation y=f(x)=3x−5y=f\left( x \right)=3x-5.
y=3x−5 ⇒3x=y+5 ⇒x=y+53 \begin{aligned} & y=3x-5 \\\ & \Rightarrow 3x=y+5 \\\ & \Rightarrow x=\dfrac{y+5}{3} \\\ \end{aligned}
Now x is expressed as a relation of y. We need to check if it can be a function or not. We need to check if it can be one to many.
Let’s assume for a fixed value c replaced in place of y, we got two solutions. This means there exists two such x for which we have one value of y.
Then applying this concept for the function y=f(x)=3x−5y=f\left( x \right)=3x-5, the function becomes two to one function. But as we already proved that the function is one-one function, we come to a contradiction.
So, we got x=y+53x=\dfrac{y+5}{3} as a function. This is the inverse function of y=f(x)=3x−5y=f\left( x \right)=3x-5.
Therefore, x=f−1(y)=y+53x={{f}^{-1}}\left( y \right)=\dfrac{y+5}{3} which gives f−1(x)=x+53{{f}^{-1}}\left( x \right)=\dfrac{x+5}{3}.

So, the correct answer is “Option B”.

Note: We need to cross-check if a relation is function or not. We have to consider that a function has an inverse only when the function is one-one. This is the condition which the inverse function also has to satisfy. In the end we just need to replace or change the variable to get the required inverse form.