## Calculus 10th Edition

Published by Brooks Cole

# Chapter 1 - Limits and Their Properties - 1.4 Exercises: 70

#### Answer

$f(g(x))$ is continuous for all values of $x$ such that $x\gt1.$

#### Work Step by Step

Using Theorem $1.12:$ $g(x)$ is continuous for all values of $x$ but $f(x)$ is continuous for $x\gt0\to$ $g(x)\gt0\to x-1\gt0\to x\gt1.$ $f(g(x))$ is continuous for all values of $x$ such that $x\gt1.$ Note: The restrictions on $f(x)$ come from the presence of a square root in the denominator; the radicand cannot be negative and the denominator cannot be $0$, hence $x\gt0.$

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