Answer
$f^{-1}(x)=\sqrt (\frac{2}{x}-1)$
Work Step by Step
$f(x)=\frac{2}{x^{2}+1}$
$y=\frac{2}{x^{2}+1}$
${x^{2}+1}=\frac{2}{y}$
${x^{2}}=\frac{2}{y}-1$
$x=\sqrt (\frac{2}{y}-1)$
$f^{-1}(x)=\sqrt (\frac{2}{x}-1)$
Calculus: Early Transcendentals (2nd Edition)
by Briggs, Bill L.; Cochran, Lyle; Gillett, Bernard
Published by
Pearson
ISBN 10:
0321947347
ISBN 13:
978-0-32194-734-5
Chapter 1 - Functions - 1.3 Inverse, Exponential, and Logarithmic Functions - 1.3 Exercises - Page 36: 28
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