Find the inverse of the function, and then determine whether the inverse is a function. 1.) y = 3 sqrt x-5 2.) y = x squared + 4

Accepted Solution

1) To find the inverse of the function [tex]y=3 \sqrt{x-5} [/tex], we are going to interchange [tex]x[/tex] and [tex]y[/tex], and then we are going to solve for [tex]x[/tex] as follows:
[tex]y=3 \sqrt{x-5} [/tex]
[tex]x=3 \sqrt{y-5} [/tex]
[tex] x^{2} =(3 \sqrt{y-5} )^{2} [/tex]
[tex] x^{2} =9(y-5)[/tex]
[tex] x^{2} =9y-45[/tex]
[tex]9y= x^{2} +45[/tex]
[tex]y= \frac{ x^{2} +45}{9} [/tex]
In this case our inverse is a parabola; since all parabolas are functions, our inverse is indeed a function.

2) To find the inverse of our function [tex]y= x^{2} +4[/tex] we are going to follow the exact same procedure as before; we'll interchange [tex]x[/tex] and [tex]y[/tex] and solve for [tex]x[/tex]:
[tex]y= x^{2} +4[/tex]
[tex]x=y^{2} +4[/tex]
[tex] y^{2} =x-4[/tex]
[tex]y=+/- \sqrt{x-4} [/tex]
Since we are taking square root, we will end with tow results: a positive and a negative one. So, our function will have tow inverses: [tex] \sqrt{x-4} [/tex] and [tex]- \sqrt{x-4} [/tex].
Notice that our original function [tex]y= x^{2} +4[/tex] is a quadratic function and by the horizontal line test (picture1), quadratic functions don't have inverse functions unless you restrict the domain. Since there is no restriction in the domain of our original function, its inverse is not a function.