Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Section 2.5 - Continuity - 2.5 Exercises - Page 125: 46

Answer

$a = \frac{1}{2}$ $b = \frac{1}{2}$

Work Step by Step

$f(x) = \frac{x^2-4}{x-2}~~~$ if $x \lt 2$ $f(x) = ax^2-bx+3~~~$ if $2 \leq x \lt 3$ $f(x) = 2x-a+b~~~$ if $x \geq 3$ For the function to be continuous at $x=2$: $\lim\limits_{x \to 2^-}\frac{x^2-4}{x-2} = f(2)$ $\lim\limits_{x \to 2^-}\frac{(x-2)(x+2)}{x-2} = a(2)^2-b(2)+3$ $\lim\limits_{x \to 2^-}(x+2) = 4a-2b+3$ $4 = 4a-2b+3$ $4a-2b-1=0$ $-8a+4b+2=0$ For the function to be continuous at $x=3$: $\lim\limits_{x \to 3^-}ax^2-bx+3 = f(3)$ $a(3)^2-b(3)+3 = 2(3)-a+b$ $9a-3b+3 = 6-a+b$ $10a-4b-3 = 0$ We can add these two equations: $-8a+4b+2=0$ $10a-4b-3 = 0$ $2a-1 = 0$ $a = \frac{1}{2}$ We can find $b$: $10a-4b-3 = 0$ $10(\frac{1}{2})-4b-3 = 0$ $5-4b-3 = 0$ $-4b = -2$ $b = \frac{1}{2}$
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