## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 2 - Limits - 2.8 Intermediate Value Theorem - Exercises - Page 86: 7

#### Answer

We have a solution $c$ in $\left[\frac{1}{4}, 2\right]$

#### Work Step by Step

Let $f(x)=\sqrt{x}+\sqrt{x+2}-3 .$ Note that $f$ is continuous on $\left[\frac{1}{4}, 2\right]$ with $f\left(\frac{1}{4}\right)=\sqrt{\frac{1}{4}}+\sqrt{\frac{9}{4}}-3=-1$ and $f(2)=\sqrt{2}-1 \approx 0.41 .$ Therefore, by the IVT, there is a $c \in\left[\frac{1}{4}, 2\right]$ such that $f(c)=\sqrt{c}+\sqrt{c+2}-3=0 .$ Thus $\sqrt{c}+\sqrt{c+2}=3$ and hence the equation $\sqrt{x}+\sqrt{x+2}=3$ has a solution $c$ in $\left[\frac{1}{4}, 2\right]$.

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