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Chapter 2: Problem 11

\(f(x)=|2 \operatorname{sgn} 2 x|+2\) \(f(0)=2\) \(\mathrm{f}\left(0^{+}\right)=4\) \(f\left(0^{-}\right)=4\) Hence, B is correct.

Short Answer

Expert verified
Question: Evaluate the function \(f(x) = |2 \operatorname{sgn} 2x| + 2\) at the points x=0, x=0^+, and x=0^-. Answer: f(0) = 2, f(0^+) = 4, and f(0^-) = 4.

Step by step solution

01

Define the given function

The given function is \(f(x) = |2 \operatorname{sgn} 2x| + 2\).
02

Understand the sign function

The sign function (sgn) is defined as follows: $\operatorname{sgn}(x) = \begin{cases} 1, & \text{if } x > 0 \\ 0, & \text{if } x = 0 \\ -1, & \text{if } x < 0 \end{cases}$
03

Evaluate f(0)

To find f(0), substitute x = 0 into the function definition: \(f(0) = |2 \operatorname{sgn}(2 \cdot 0)| + 2 = |2 \operatorname{sgn}(0)| + 2 = |2 \cdot 0| + 2 = 2\).
04

Evaluate f(0^+)

To find f(0^+), evaluate the function at a point slightly greater than 0: \(f(0^+) = |2 \operatorname{sgn}(2 \cdot 0^+)| + 2 = |2 \operatorname{sgn}(0^+)| + 2 = |2 \cdot 1| + 2 = 4\).
05

Evaluate f(0^-)

To find f(0^-), evaluate the function at a point slightly less than 0: \(f(0^-) = |2 \operatorname{sgn}(2 \cdot 0^-)| + 2 = |2 \operatorname{sgn}(0^-)| + 2 = |2 \cdot (-1)| + 2 = 4\). Now, we can see that the correct answer is B, since f(0) = 2, f(0^+) = 4, and f(0^-) = 4.

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Most popular questions from this chapter

$$ \lim _{x \rightarrow 1^{\prime}} F(x)=\lim _{x \rightarrow 1^{*}} \operatorname{sgn}(x-1) \cot ^{-1}[x-1] $$ $$ \begin{aligned} &=\frac{\pi}{2} \\ \lim _{x \rightarrow 1^{-}} F(x) &=\lim _{\left.x \rightarrow\right|^{-}} \operatorname{sgn}(x-1) \cot ^{-1}[x-1] \\ &=-\frac{3 \pi}{4} \end{aligned} $$

$\lim _{x \rightarrow 0^{\prime}} \cos \left(x \cos \frac{1}{x}\right)=1 \quad f(0)$ is not defined. \(\lim _{x \rightarrow 0^{-}} \cos \left(x \cos \frac{1}{x}\right)=1\)

\(2 \leq \frac{f(x)}{x^{2}} \leq 3\) \(2 x^{2} \leq f(x) \leq 3 x^{2}\) If \(x=\frac{1}{2}\) $$ \rightarrow \frac{1}{2} \leq f\left(\frac{1}{2}\right) \leq \frac{3}{4} $$ If \(x=1\) If \(x=\frac{1}{3}\) \(\quad \rightarrow 2 \leq f(1) \leq 3\) $\quad \rightarrow \frac{2}{9} \leq f\left(\frac{1}{3}\right) \leq \frac{1}{3}$ $\begin{aligned} \mathrm{f}_{\mathrm{x}}=\frac{1}{4} & \\\$$$ $$$& \rightarrow \frac{1}{8} \leq \mathrm{f}\left(\frac{1}{4}\right) \leq \frac{3}{16} \\ \mathrm{f} x=\frac{1}{8} & \\ & \rightarrow \frac{1}{32} \leq \mathrm{f}\left(\frac{1}{8}\right) \leq \frac{3}{64} \end{aligned}$

$\lim _{x \rightarrow 0} \frac{\mathrm{e}^{x}-\mathrm{e}^{[x]}}{\mathrm{e}^{x}}=1-\mathrm{e}$ \(\lim _{x \rightarrow 0^{-}} \frac{\sin \\{x\\}}{\\{\tan x\\}}=1\) $\lim _{x \rightarrow \frac{\pi}{6}} \frac{\sin \\{x\\}}{\\{\tan x\\}}=\frac{\sin \pi / 6}{\tan \pi / 6}=\frac{\sqrt{3}}{2}$ $\lim _{x \rightarrow \frac{\pi}{6}} \frac{\sin \\{x\\}}{\\{\tan x\\}}=\frac{\sqrt{3}}{2}, f\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}$ $\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin \\{x\\}}{\\{\tan x\\}}=\frac{\sin \pi / 4}{\tan \frac{\pi}{4}-1}=\infty$

\(\lim _{x \rightarrow 0^{\prime}} \frac{b e^{x}-\cos x-x}{x^{2}}\) For limit to exist, \(b=1\) \(\lim _{x \rightarrow 0^{+}} \frac{e^{x}-\cos x-x}{x^{2}}=1=a\) $\lim _{x \rightarrow 0^{-}} \frac{2\left(\tan ^{+} e^{x}-\frac{\pi}{4}\right)}{x}=1$
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