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Question: If we have a function as \(f\left( x \right) = 3x - 1\) and if \({f^{ - 1}}\) is the inverse functio...

If we have a function as f(x)=3x−1f\left( x \right) = 3x - 1 and if f−1{f^{ - 1}} is the inverse function of f, then what is f−1(5){f^{ - 1}}\left( 5 \right) ?
(A) 1818
(B) 43\dfrac{4}{3}
(C) 53\dfrac{5}{3}
(D) 22
(E) 00

Explanation

Solution

In the problem, we are required to first find the inverse function of the function given to us. Then, we substitute the value of the variable as specified in the question to get to the final answer. This question requires us to have the knowledge of basic and simple algebraic rules and operations such as substitution, addition, multiplication, subtraction and many more like these. A thorough understanding of functions and its inverse can be of great significance.

Complete step-by-step solution:
Here, f(x)=3x−1f\left( x \right) = 3x - 1 and f−1{f^{ - 1}} is the inverse function.
So, we first assume the function to be the variable y. So, we get,
f(x)=y=3x−1f\left( x \right) = y = 3x - 1
Now, to find the inverse of the function, we have to find the value of x in terms of y. So, we get,
⇒y=3x−1\Rightarrow y = 3x - 1
Adding 11 to both sides of the equation, we get,
⇒y+1=3x−1+1\Rightarrow y + 1 = 3x - 1 + 1
Simplifying the equation and dividing both sides by 33, we get,
⇒y+1=3x\Rightarrow y + 1 = 3x
⇒y+13=3x3\Rightarrow \dfrac{{y + 1}}{3} = \dfrac{{3x}}{3}
Carrying out the calculations, we get the value of x in terms of y as,
⇒x=y+13\Rightarrow x = \dfrac{{y + 1}}{3}
Hence, we get the inverse function as f−1(y)=y+13{f^{ - 1}}\left( y \right) = \dfrac{{y + 1}}{3}.
Now, we have to find the value of the inverse function for y=5y = 5.
So, replacing y with five in the inverse function, we get,
f−1(5)=5+13{f^{ - 1}}\left( 5 \right) = \dfrac{{5 + 1}}{3}
Carrying out the calculations, we get,
⇒f−1(5)=63=2\Rightarrow {f^{ - 1}}\left( 5 \right) = \dfrac{6}{3} = 2
Therefore, option (D) is the correct answer.

Note: Such questions that require just a simple change of variable can be solved easily by keeping in mind the algebraic rules such as substitution and transposition. Substitution of a variable involves putting a certain value in place of the variable. That specified value may be a certain number or even any other variable.