University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.6 - Inverse Functions and Logarithms - Exercises - Page 49: 40

Answer

(a) $$f^{-1}(x)=-x+1$$ The graphs are shown in the image below. The line $y=-x+1$ and the line $y=x$ intersect at $90^\circ$. (b) $$f^{-1}(x)=-x+b$$ The line $y=-x+b$ makes with the line $y=x$ an angle of $90^\circ$. (c) The inverses of those functions, whose graphs are perpendicular to the line $y=x$, are themselves.
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Work Step by Step

(a) $$y=f(x)=-x+1\hspace{1cm}$$ - To find its inverse: 1) Solve for $x$ in terms of $y$: $$y=-x+1$$ $$x=-y+1$$ 2) Interchange $x$ and $y$: $$y=-x+1$$ Therefore, $$f^{-1}(x)=-x+1$$ The graphs are shown in the image below. From the graphs, we can see that the line $y=-x+1$ and the line $y=x$ intersect at $90^\circ$. (b) $$y=f(x)=-x+b\hspace{1cm}$$ - To find its inverse: 1) Solve for $x$ in terms of $y$: $$y=-x+b$$ $$x=-y+b$$ 2) Interchange $x$ and $y$: $$y=-x+b$$ Therefore, $$f^{-1}(x)=-x+b$$ We can deduce from (a), or notice that the slope of $y=-x+b$ is $-1$ and the slope of $y=x$ is $1$ and since $-1\times1=-1$, these two lines would make an angle of $90^\circ$ with each other. (c) From (a) and (b), we notice that the formula of the inverse are exactly the same as the formula of the original functions. That means, their graphs must overlap each other, or in other words, the inverse of those functions, whose graphs are lines perpendicular to the line $y=x$, are themselves.
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