Question

How would you do the inverse of this function: -1/7sqrt(16-x^2)?

-I know you switch the x and y and solve for y.

I keep on getting a weird answer. Help!

Answer 1

What did you get for f^-1(x)?

And what did you get for f^-1(f(x))?

Answer 2

In general, there are different ways you can verify your answer, and in this case, first verify the question!

For a function f(x) to have an inverse, it must be one-to-one and onto.

A function must pass the vertical line test (i.e. <i>any</i> vertical line within the domain must not have two or more values for the image).

A one-to-one function must pass the vertical line test <i>and</i> the horizontal line test (i.e. <i>any</i> horizontal line within the range must not have two or more values in the domain).

The given function does <i>not</i> pass the horizontal line test because it is an even function, i.e. for any value of x on its domain (-4,4), f(x)=f(-x).

Before finding the inverse, we must restrict the domain of the given function to either [0,4) or (-4,0]. We will choose [0,4). The other option can be treated similarly.

If you work out the math, which I am sure you did, after the solution of the equations, the inverse turns out to be either
g(x)=-sqrt(784*x^2-1)/(7*x) or
g(x)=sqrt(784*x^2-1)/(7*x)

Which one do we choose?

The answer lies in the domain that we have chosen in the first place.

We have chosen the domain to be [0,4), for which the image is always negative. Thus the image of the inverse g(x) has to be [0,4), and its domain has to be negative.

For g(x) to be positive when the domain is negative, we can only choose the first option, or
g(x)=-sqrt(784*x^2-1)/(7*x)

A graphics plot will make this all clear:

http://imageshack.us/photo/my-images/43/1312958033.png/

Answer 3

On re-reading the question, I just realized from the way you have punctuated the question, f(x) could be either
f(x)=-1/7sqrt(16-x^2)
or simply
f(x)=1/7sqrt(16-x^2).

In the above response, I have taken it to be
f(x)=-1/7sqrt(16-x^2)

If f(x) has the opposite sign, it will be reflected about the x-axis, and the inverse about the y-axis.
It would make a good exercise for your arguments in this case.