Question

How do you find the inverse of $Y = 5X - 1$ and is it a function ?

Answer 1

Hint: In this question, we need to find the inverse function of a given function. To find the inverse function, we need to exchange the variables x and y with each other. We further make simplifications and to solve for y, we isolate it to get our new inverse function in terms of the variable x. After that we verify, whether it is a function or not.

Complete step by step answer:
Given the function of the form $Y = 5X - 1$ …… (1)
We are asked to determine the inverse function of the above function and also check whether it is also a function or not.
Let us now switch between the variables x and y.
i.e. we will exchange the variables x and y with each other in the equation (1), we get,
$ \Rightarrow X = 5Y - 1$
We will rewrite it as,
 $ \Rightarrow 5Y - 1 = X$
Adding 1 on both sides of the equation above, we get,
$ \Rightarrow 5Y - 1 + 1 = X + 1$
Combining like terms $ - 1 + 1 = 0$
Hence we have,
$ \Rightarrow 5Y + 0 = X + 1$
$ \Rightarrow 5Y = X + 1$
Divide by 5 on both sides of the equation, we get,
$ \Rightarrow \dfrac{{5Y}}{5} = \dfrac{{X + 1}}{5}$
Cancelling the common terms, we get,
$ \Rightarrow Y = \dfrac{{X + 1}}{5}$
After rearranging we get,
$ \Rightarrow Y = \dfrac{X}{5} + \dfrac{1}{5}$
This is the required inverse function.
Note that it is also a function. Because one output for each input value of $X$ is obtained by the above resultant equation.

Hence, the inverse of the function $Y = 5X - 1$ is given by $Y = \dfrac{X}{5} + \dfrac{1}{5}$ and it is also a function.

Note: Note that finding the inverse function of a linear equation is relatively easier than finding the inverse of a quadratic or cubic equation. This is because the domain and range of the linear equations span over all the real numbers, given that the domain is not restricted.
Remember that to find the inverse function for a linear equation, we need to exchange the variables given with each other and then simplify it further.
Not all quadratic equations have inverses. We can find the inverse only if the range and the domain are well defined.