Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Section 3.10 - Linear Approximations and Differentials. - 3.9 Exercises - Page 257: 44


(a) $g(1.95) \approx -4.15$ $g(2.05) \approx -3.85$ (b) The approximations found in part (a) are too small.

Work Step by Step

(a) When $x = 2$: $g(2) = -4$ $g'(2) = \sqrt{x^2+5} = \sqrt{2^2+5} = 3$ We can find the linear approximation at $a=2$: $g(x) \approx g(a)+g'(a)(x-a)$ $g(x) \approx g(2)+g'(2)(x-2)$ $g(x) \approx (-4)+3(x-2)$ $g(x) \approx 3x-10$ We can approximate $g(1.95)$: $g(x) \approx 3x-10$ $g(1.95) \approx 3(1.95)-10$ $g(1.95) \approx -4.15$ We can approximate $g(2.05)$: $g(x) \approx 3x-10$ $g(2.05) \approx 3(2.05)-10$ $g(2.05) \approx -3.85$ (b) As $x$ increases, the value of $g'(x)$ increases. Therefore, $g(x)$ is concave up. If a line is tangent to a function that is concave up, then the tangent line is below the function, so the approximations found in part (a) are too small.
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