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Question: How do you graph \(f\left( x \right)=2x+1\) and then use the horizontal test to determine whether th...

How do you graph f(x)=2x+1f\left( x \right)=2x+1 and then use the horizontal test to determine whether the inverse of f(x)f\left( x \right) is a function?

Explanation

Solution

We first explain the relation of inverse function with its function types. We find if the function is one-one or not. We use horizontal line tests to determine whether the inverse of f(x)f\left( x \right) is a function and find the function if it exists.

Complete step-by-step solution:
We need to find whether the function f(x)=2x+1f\left( x \right)=2x+1 has an inverse or not through a horizontal line test. If it has an inverse then we find the inverse.
First, we find the characteristics of the function f(x)=2x+1f\left( x \right)=2x+1. It is a one-one function.
This means for a particular value of xx, we will get only one value of yy.
So, every value of xx can be projected to a particular value of yy.
In case of one-one function, we can have an inverse always irrespective of the function.
In case many-one functions, we can’t have an inverse function.
Let’s assume that g(a)=g(b)=cg\left( a \right)=g\left( b \right)=c for an arbitrary function g(x)g\left( x \right).
This is a two-one function. The domain has two values a,ba,b that give the same value of cc in range. Condition is aba\ne b.
When we are taking the inverse, we get one value cc in domain which gives two values a,ba,b in range. That can’t be function as g1(c)=a{{g}^{-1}}\left( c \right)=a and g1(c)=b{{g}^{-1}}\left( c \right)=b but aba\ne b.
The horizontal line test gives that for the function f(x)f\left( x \right) if any horizontal line represented as y=ky=k on infinite extension cuts the graph more than once then the graph can’t have an inverse function.
For f(x)=2x+1f\left( x \right)=2x+1, we have inverse function. Let y=2x+1y=2x+1 which gives x=y12x=\dfrac{y-1}{2}.
Therefore, the inverse function is f1(x)=x12{{f}^{-1}}\left( x \right)=\dfrac{x-1}{2}.
We take a horizontal line represented as y=2y=2. We draw the graphs of f(x)=2x+1f\left( x \right)=2x+1 and y=2y=2.
We find their intersection.

The only intersection is the point A=(12,2)A=\left( \dfrac{1}{2},2 \right). Therefore, the inverse of f(x)f\left( x \right) exists.

Note: If the horizontal line intersects the graph of a function in all places at exactly one point, then the given function should have an inverse that is also a function. We say this function passes the horizontal line test.