Finding the domain of f-1(x).

Question

Given f(x) = 2/3 X² - 7. Find the domain of f^-1(x).

Answer 1

If the function is f(x) = (2/3)x² - 7, then it does not have an inverse function because the function is not one-to-one (in other words, the graph fails the horizontal line test).  

 

On the other hand, if we consider the function F(x) = (2/3)x² - 7, where x ≥ 0, then the function is one-to-one and an inverse function exists.  The domain of F(x) is [0, ∞) and the range is [-7, ∞).  Since the domain of F^-1(x) is the range of F(x), the domain of F^-1(x) is [-7, ∞).  

Answer 2

The formatting of f(x) makes it hard to determine which function we are trying to invert.

Are we considering f(x)= 2/(3x^2-7) ??

If so, this function in general doesn't have an inverse, as f(x)=f(-x), because the squared power in the only instance of x, 'kills' the negative sign. This means the function we are considering is EVEN. And for every value for x we have in our DOMAIN, there is a second value (namely the negative) that produces the same output. This means that the function is necessarily not one-to-one (i.e. it graphically fails the horizontal line test) and therefor cannot have a well defined inverse!

In order to resolve an issue like this and consider f^{-1}, we need to restrict the DOMAIN of f to a subset of the real line, so that it is 1-1 and onto that subDOMAIN and corresponding IMAGE of the new restricted subDOMAIN, and will thus have an inverse as a restricted function.

 

First lets identify the domain of f...the denominator cannot be 0, so 3x^2-7=0 implies x=+/- \sqrt{7/3}...so our DOMAIN is all real numbers except these two values.  We can then restrict f to the subDOMAIN [0,\sqrt{7/3}) U (\sqrt{7/3}, \infinity).  We can check that this function f is 1-1 on this subDOMAIN, that is that f(x)=f(y) implies x=y...each input gives a unique output!

 

Next we need to find the IMAGE/RANGE of this function on this subDOMAIN.  Note that when values for x close to \sqrt{7/3} are input into f, we get outputs going off to +/- infinity!  In general for an arbitrary real number y we can find a value for x, such that f(x)=y by solving the equation y=2/(3x^2-7) for x:

 

x =\sqrt{(2/y+7)/3}=\sqrt{2+7y)/(3y)} = f^{-1}(y).

 

We then note that it is not possible for f(x) to be equal to 0, since y=0 is not a valid option for this inverse functions input.  Thus the RANGE/IMAGE of f on our restricted subDOMAIN is (-\infinity,0) U (0,\infinity).  This is the domain of the inverse function we have constructed above.

 

I hope this helps...