Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 1 - Section 1.5 - Inverse Functions and Logarithms - 1.5 Exercises: 31

Answer

a) $f^{-1}(x)=\sqrt{1-x^2}=f(x)$ The inverse is the same as the original function. b) The graph of $f$ is a quarter of a circle with centre at the origin and radius $1$(u) located in the first quadrant. Since $f$ is symmetrical about $y=x$, its inverse is identical to itself.

Work Step by Step

a) $f(x)=y=\sqrt{1-x^2}, 0\leq x\leq1$ $y^2=1-x^2$ $x^2=1-y^2$ $x=\pm\sqrt{1-y^2}\geq0$ $\therefore x=\sqrt{1-y^2}$ $y\rightleftharpoons x$ $f^{-1}(x)=y=\sqrt{1-x^2}=f(x)$ The inverse is the same as the original function. b) $f(x)=y=\sqrt{1-x^2}$ Domain: $0\leq x\leq1$ Range: $0\leq y\leq1$ $\therefore$Graph exists only in 1st quadrant Squaring both sides: $y^2=1-x^2$ $(x-0)^2+(y-0)^2=1^2$ Circle with centre $(0,0)$, r=$1$(u) The graph of $f$ is a quarter of a circle with centre at the origin and radius $1$(u) located in the first quadrant. Since $f$ is symmetrical about $y=x$, its inverse is identical to itself.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.