Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.1 Inverse Functions - 6.1 Exercises: 33

Answer

(a) the function $f(x)=\sqrt {1-x^{2}}$ and $f^{-1}(x)=\sqrt {1-x^{2}}$ is same. (b) Both the functions $f(x) and $f^{-1}(x)$ represent the same graph which forms an arc of a unit circle in the positive quadrant as depicted below:
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Work Step by Step

(a) Calculate the inverse of the function $f(x)=\sqrt {1-x^{2}};0\leq x\leq 1$ Write $y=f(x)$ $y=\sqrt {1-x^{2}}$ Solve this equation for x in terms of y to get the inverse function. $y^{2}=1-x^{2}$ $x^{2} + y^{2} =1$ $x=\sqrt {1-y^{2}}$ To express $f^{-1}(x)$ as a function of x,interchange x and y. The resulting equation is $y=\sqrt {1-x^{2}}$ Therefore, the inverse of the function $f^{-1}(x)=y=\sqrt {1-x^{2}}$ Hence, the function $f(x)=\sqrt {1-x^{2}}$ and $f^{-1}(x)=\sqrt {1-x^{2}}$ is same. (b) Solve this equation for x in terms of y to get the inverse function. $y^{2}=1-x^{2}$ $x^{2} + y^{2} =1$ Both the functions $f(x) and $f^{-1}(x)$ represent the same graph which forms an arc of a unit circle in the positive quadrant as depicted below:
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