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

 Aug 3, 2022
 #1
avatar+113 
+2

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

I think this looks a little better:

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

Let's first find f(-2):

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

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):

 

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

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...

 Aug 3, 2022
 #2
avatar+33151 
+2

I get the following:

 

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

 Aug 3, 2022
 #3
avatar+113 
+2

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

tuffla2022  Aug 3, 2022
 #4
avatar+33151 
0

No worries - we all make them!

Alan  Aug 3, 2022

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