Chapter 2.4, Problem 54E

a.

To determine

To prove:that the given absolute value function $F\left(x\right)=\left|x\right|$ is continuous are positive equation.

Explanation of Solution

Given:

The givenabsolute function is $F\left(x\right)=\{\begin{array}{l}xifx>0\\ -xifx<0\end{array}$ .

Calculation:

The given function is $F\left(x\right)=\{\begin{array}{l}xifx>0\\ -xifx<0\end{array}$ .

It is clear that $F\left(x\right)$ are continuous on $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ . Since, polynomial functions are continuous.All we have left to do is show that F(x) is continuous at x = 0. We need to show that the left sided limit and the right sided limit at x = 0 are equal.

As $x$ approaches 0 from the negative direction, then $x<0,\text{so }F\left(x\right)=-x$

Therefore, by squeeze theorem, $\underset{x\to 0^{-}}{\lim }F\left(x\right)=\underset{x\to 0^{-}}{\lim }\left(-x\right)=-0=0$ .

As $x$ approaches 0from the positive direction, then $x>0,\text{so }F\left(x\right)=x$

Therefore, by squeeze theorem, $\underset{x\to 0^{+}}{\lim }F\left(x\right)=\underset{x\to 0^{+}}{\lim }\left(x\right)=0$ .

Since the left sided limit is equal to the right sided limit.

Therefore, $\underset{x\to 0}{\lim }F\left(x\right)=0$

Since the absolute value of 0 equals 0.

Therefore, $F\left(0\right)=0$

Thus, $F\left(x\right)$ is continuous at $x=0$

Therefore, the given function $F\left(x\right)$ is continuous on $\left(-\infty ,\infty \right)$ .

b.

To determine

To prove: that the given absolute value function $F\left(x\right)=\left|x\right|$ is continuous on $I$ .

Explanation of Solution

Given:

Let f be continuous on a given interval.

Calculation:

Let f be continuous on a given interval.

From part (a) the given function $F\left(x\right)$ is continuous on $\left(-\infty ,\infty \right)$ .

The absolute value function is continuous on $I$

Since, by theorem 9 - the composite of continuous functions is also a continuous function.

Therefore, $F\left(x\right)$ is continuous on $I$ .

c.

To determine

To check: that the converse of statement in part(b) is also true. If not then find a counter example.

Explanation of Solution

Given:

From part (b) $F\left(x\right)$ is continuous on $I$ .

Calculation:

The converse of statement in part(b) is not true.

Now, it is to show that $\left|f\left(x\right)\right|$ is continuous while $f\left(x\right)$ is not.

Counterexample:

Let $f\left(x\right)={\left(-1\right)}^x$ and the intervalis $\left(-\infty ,\infty \right)$ .

Therefore, $f\left(x\right)={\left(-1\right)}^x=1$ , which is continuous for all x in $\left(-\infty ,\infty \right)$ .

  $f\left(x\right)={\left(-1\right)}^x$ is undefined for fractions with even denominators. So, clearly, f is not continuous at these points. Therefore, $F\left(x\right)$ is not continuous on $\left(-\infty ,\infty \right)$ .