Inverse Functions

[tchaikovsky]

For each function f, find f^-1 and the domain and range of f and f^-1. Determine whether f^-1 is a function.

f(x) = 2x^2 + 2

So far I have:
f(x) = 2x^2 + 2
x = 2y^2 + 2
x - 2 = 2y^2
1/2x - 1 = y^2
+/- sqrt(1/2 - 1) = y

So f^-1(x) = +/- sqrt(1/2x - 1) ????

I'm not sure about the domain and range.

 

[stapel]

What is the domain of the original function f(x)? What is its range? (You can find the range from the graph, of course.)

By nature of inverses, the domain and range of the inverse relation are the reverse of the domain and range of the original relation. So just reverse the answers from above.

I agree with your expression for the inverse relation, which I take to be the following:


. . . . .\(\displaystyle \large{y\,=\,\pm\,\sqrt{\frac{1}{2}x\,-\,1}}\)


But how do you know this to be a function?

Eliz.

 

[TchrQbic]

For each function f, find f^-1 and the domain and range of f and f^-1. Determine whether f^-1 is a function.

f(x) = 2x^2 + 2

So far I have:
f(x) = 2x^2 + 2
x = 2y^2 + 2
x - 2 = 2y^2
1/2x - 1 = y^2
+/- sqrt(1/2 - 1) = y

So f^-1(x) = +/- sqrt(1/2x - 1) ????

I'm not sure about the domain and range.

Remember that the domain of a function includes all possible real values for the variable except, in this case, those values that would make the radicand negative.

To check if it's a function, test values of the independent variable to see if you get two resulting values. If one x-value pairs with two y-values, it is NOT a function.

 

[Gene]

Does that mean you have the answers or is something still open?
---------------
Gene

 

[tchaikovsky]

Thanks a lot, I'm pretty sure I've got it now.