Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.1 Real Numbers, Functions, and Graphs - Exercises - Page 10: 26


The solution is $(-\infty, 2 ]$.

Work Step by Step

Recall that: the square root of $k^2$ is $|k|$ (for all real numbers $k$). Thus, the problem could be rewritten as $\sqrt{(x-3)^2} = \sqrt{(x-2)^2} + 1$ by squaring the two sides, we get $(x^2 -6x+9) = (x^2 -4x +4) +(1) + 2 |x-2|$ Combining like terms gives: $-2x+4= -2(x-2) = 2 |x-2|$ Thus, we have $-(x-2)= |x-2|$ From the definition of the absolute value function, we get $x-2 \le 0$, and then $x \le 2$. Hence, the solution as a half infinite interval is $(-\infty, 2]$.
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