Given that f(x) = x^{-1} + \frac{x^{-1}}{1+x^{-2}}, what is f(f(-2))? Express your answer as a common fraction.

Guest Aug 3, 2022

#1**+2 **## \(f(x) = x^{-1} + \frac{x^{-1}}{1+x^{-2}}\)

## \(f(x) = \frac{1}{x} + \frac{\frac{1}{x}}{1+\frac{1}{x^2}}\)

## \(f(-2) = \frac{1}{-2} + \frac{\frac{1}{-2}}{1+\frac{1}{(-2)^2}}\)

# \(f(\frac{-9}{10}) = \frac{1}{\frac{-9}{10}} + \frac{\frac{1}{\frac{-9}{10}}}{1+\frac{1}{(\frac{-9}{10})^2}}\)

I think this looks a little better:

Let's first find f(-2):

we can simplify this somewhat, since the denominator in the second term is

1 + 1/4 = 5/4, and so the second term becomes - 4/10 or -2/5

f(-2) = -1/2 - 2/5 = -5/10 - 4/10 = - 9/10

Next we must find f(-9/10):

The first term evaluates to -10/9, ans so does the numerator in the second term.

The denominator is 1 + 1/(81/100), or 1 + 100/81, that is 181/81.

So the second term now evaluates to (-10/9)/(181/100) = -1,000/729

The sum of the two terms is -10/9 - 1,000/729 = -810/729 - 1,000/729 = -1,810/729 or -2 -352/729.

That is, if my calculations are correct... This was quite an excercise, so maybe I'm wrong somewhere...

tuffla2022 Aug 3, 2022

#2**+2 **

I get the following:

(Check your calculation "*So the second term now evaluates to (-10/9)/(181/100) = -1,000/729*" tuffla2022)

Alan Aug 3, 2022

#3**+2 **

Thank You, Alan. It was late, and I must have made some simple error somewhere.

tuffla2022
Aug 3, 2022