Answer
(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.
Work Step by Step
(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.
