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What is the inverse of f(x) = (x^2 +1) / (1- x^2)?

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lxsptter | Student, Undergraduate

Posted October 20, 2010 at 5:05 PM via web

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What is the inverse of f(x) = (x^2 +1) / (1- x^2)?

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william1941 | College Teacher

Posted October 20, 2010 at 5:08 PM (Answer #1)

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The inverse function of f(x) is defined as the function which when applied to the result of f(x) gives x.

We have f(x) = (x^2 +1) / (1- x^2), let this be equal to y

=> y = (x^2 +1) / (1- x^2)

=> y ( 1-x^2) = x^2 +1

=> y – y x^2 = x^2 +1

=> y -1 = yx^2 + x^2

=> x^2( 1+y) = y-1

=> x^2 = (y-1) / (y+1)

=> x = sqrt [(y-1) / (y+1)]

Interchange y and x

y = sqrt [(x-1) / (x+1)]

f^-1(x) = sqrt [(x-1) / (x+1)]

Therefore the inverse function of f(x) = (x^2 +1) / (1- x^2) is f^-1(x) = sqrt [(x-1) / (x+1)]

To verify: f(x) = (x^2 +1) / (1- x^2)

substitute this in sqrt [(x-1) / (x+1)]

=> sqrt [{(x^2 +1) / (1- x^2) -1} / { (x^2 +1) / (1- x^2)} +1]

=> sqrt[ (x^2 +1 – 1 + x^2) / x^2 + 1 + 1 – x^2]

=> sqrt [ 2x^2 /2]

=> sqrt x^2

=> x

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neela | High School Teacher

Posted October 20, 2010 at 7:46 PM (Answer #2)

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To find the inverse of f(x) = (x^2+1)/(1-x^2).

This is not a bijection unless we ristrict x only for positve value or x for only negative value so that the inverse function can exist.

Let  y = f^-1(x).

Then f(y) = x.

x =f(y) = (y^2+1)/(1-y^2)

x(1-y^2) = y^2+1

x-1 = xy^2+y^2

x-1 = (x+1)y^2

y ^2 = (x-1)/(x+1)

y = sqrt{(x-1)/(x+1)} , x > 1.

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