University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 4 - Section 4.2 - The Mean Value Theorem - Exercises - Page 223: 23

Answer

See proof below.

Work Step by Step

$g(t)=\sqrt{t}+\sqrt{1+t}-4=t^{1/2}+(1+t)^{1/2}-4$ is continuous and differentiable on $(0,\infty)$. $g(1)\lt 0, \qquad g(16)\gt 0$ so the Intermediate Value Theorem guarantees that there is a zero between $1$ and $16$. Check whether $g$ is increasing over $(0,\infty)$. (If it is, the mentioned zero would be the only zero.) $g'(t)=\displaystyle \frac{1}{2}(t^{-1/2}-(1+t)^{-1/2})$ $t^{-1/2}=\displaystyle \frac{1}{\sqrt{t}}$ is a positive number, and $(1+t)^{-1/2}=\displaystyle \frac{1}{\sqrt{t+1}} $has a greater denominator than $\displaystyle \frac{1}{\sqrt{t}},$ so $t^{-1/2}-(1+t)^{-1/2}\gt 0$. So, for $ t\in (0,\infty),\qquad g'(t)\gt 0$ and $g$ is an increasing function on $(0,\infty)$ So, $g$ has only one zero in $(0,\infty)$
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