Calculus: Early Transcendentals (2nd Edition)

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: 36


$f(x)=\dfrac{6}{x^{2}-9}$ $f^{-1}(x)=\sqrt{9+\dfrac{6}{x}}$

Work Step by Step

$f(x)=\dfrac{6}{x^{2}-9}$, for $x\gt3$ Substitute $f(x)$ by $y$: $y=\dfrac{6}{x^{2}-9}$ Solve the equation for $x$. Start by taking $x^{2}-9$ to multiply the left side: $y(x^{2}-9)=6$ Take $y$ to divide the right side: $x^{2}-9=\dfrac{6}{y}$ Take $9$ to the right side: $x^{2}=9+\dfrac{6}{y}$ Finally, take the square root of both sides: $\sqrt{x^{2}}=\sqrt{9+\dfrac{6}{y}}$ $x=\sqrt{9+\dfrac{6}{y}}$ Interchange $x$ and $y$: $y=\sqrt{9+\dfrac{6}{x}}$ Substitute $y$ by $f^{-1}(x)$: $f^{-1}(x)=\sqrt{9+\dfrac{6}{x}}$ The graph of both the original function and its inverse is shown in the answer section.
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