## Calculus: Early Transcendentals 8th Edition

(a) $F(x) = \vert x \vert$ is continuous for all real numbers. (b) If $f$ is a continuous function on an interval, then so is $\vert f \vert$ (c) The converse of the statement in part (b) is not true. If $\vert f \vert$ is continuous, it does not follow in general that $f$ is continuous.
(a) Choose any real number $c$. We will show that the function $F(x) = \vert x \vert$ is continuous at $c$ Let $\epsilon \gt 0$ be given. Let $\delta = \epsilon$ Suppose that $0 \lt \vert x-c \vert \lt \delta$ Then: $\vert F(x)-F(c) \vert = \vert ~~\vert x \vert-\vert c \vert ~~\vert \leq \vert x-c \vert \lt \delta = \epsilon$ Therefore, $F(x)$ is continuous at $c$. Since $c$ was chosen arbitrarily, $F(x)$ is continuous for all real numbers. (b) Suppose $f$ is continuous on an interval. Choose any number $c$ in this interval. Let $\epsilon \gt 0$ be given. There is a positive number $\delta$ such that if $0 \lt \vert x-c \vert \lt \delta$ then $\vert f(x)-f(c) \vert \lt \epsilon$ Then: $\vert ~~\vert f(x) \vert-\vert f(c) \vert ~~\vert \leq \vert f(x)-f(c) \vert \lt \epsilon$ Therefore, $\vert f \vert$ is continuous at $c$. Since $c$ was chosen arbitrarily, $\vert f \vert$ is continuous for all numbers in the interval. (c) The converse of the statement in part (b) is not true. If $\vert f \vert$ is continuous, it does not follow in general that $f$ is continuous. We can verify this with the following counterexample. $f(x) = 1~~~$ if $x$ is a rational number $f(x) = -1~~~$ if $x$ is an irrational number Clearly $f$ is not continuous. However, $\vert f \vert = 1$ for all real numbers and clearly $\vert f \vert$ is continuous. Note that $\vert f \vert$ is continuous, but $f$ is not continuous.