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Question

Question: How do you find the equation for the inverse of \(y=2x+1\) ?...

How do you find the equation for the inverse of y=2x+1y=2x+1 ?

Explanation

Solution

For answering this question we need to find the inverse equation of the given function. The inverse equation of the given function is given by interchanging the domain of the function with the range. For a function the domain is xx and range is yy and for inverse the domain is yy and the range is xx . So let us find the value of xx in terms of yy .

Complete answer:
Now considering from the question we need to find the equation for the inverse of the given function y=2x+1y=2x+1 .
From the basic concepts we know that the inverse equation of the given function is given by interchanging the domain of the function with the range. For a function the domain is xx and range is yy and for inverse the domain is yy and the range is xx .
Now we need to find the value of xx in terms of yy .
By shifting 11 from left hand side to right hand side we will have y1=2x\Rightarrow y-1=2x .
By dividing the whole equation by 22 on both sides we will have the value of xx .
After doing the simplification we will have y12=x\Rightarrow \dfrac{y-1}{2}=x .
After doing all these simplifications we now have a reduced equation giving the value of xx in terms of yy .
To get the inverse equation we need to interchange xx and yy.
After doing this we will have y=x12y=\dfrac{x-1}{2} .
Therefore we can conclude that the equation for the inverse of the given function y=2x+1y=2x+1 is given as y=x12y=\dfrac{x-1}{2}

Note: While answering this question we need to make sure of the calculations and the basic concepts of inverse equations. We should surely interchange xx and yy for the inverse equation otherwise the equation we obtained will not be the inverse it will remain the same. Similarly we can give the inverse of any function but we should make sure with some aspects like dividing with zero is not defined so for the function xy=1xy=1 the inverse of the function will be y=1xy=\dfrac{1}{x} but for x=0x=0 the inverse function will not be defined.