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 - Page 36: 37


$f(x)=x^{2}-2x+6$ $f^{-1}(x)=1+\sqrt{x-5}$

Work Step by Step

$f(x)=x^{2}-2x+6$, for $x\ge1$ Substitute $f(x)$ by $y$: $y=x^{2}-2x+6$ Solve the equation for $x$. Start by completing the square on the right side. $y=(x^{2}-2x)+6$ Complete the square by adding $\Big(\dfrac{b}{2}\Big)^{2}$ inside the parentheses and subtracting it outside of it. In this case, $b=-2$ $y=\Big[x^{2}-2x+\Big(-\dfrac{2}{2}\Big)^{2}\Big]+6-\Big(-\dfrac{2}{2}\Big)^{2}$ $y=(x^{2}-2x+1)+6-1$ $y=(x^{2}-2x+1)+5$ Factor the expression inside the parentheses, which is a perfect square trinomial: $y=(x-1)^{2}+5$ Take $5$ to the left side: $y-5=(x-1)^{2}$ Take the square root of both sides: $\sqrt{y-5}=\sqrt{(x-1)^{2}}$ $\sqrt{y-5}=x-1$ Take $1$ to the left side and rearrange the equation: $x=1+\sqrt{y-5}$ Interchange $x$ and $y$: $y=1+\sqrt{x-5}$ Substitute $y$ by $f^{-1}(x)$: $f^{-1}(x)=1+\sqrt{x-5}$ The graph of both the original function and its inverse is shown in the answer section.
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