/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 87 $f(x)=[x] \& g(x)= \begin{ca... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 87

$f(x)=[x] \& g(x)= \begin{cases}0, & x \in I \\ x^{2}, & \text { otherwise }\end{cases}$ \(g \circ f(x)=0, \quad x \in R\) $f \circ g(x)= \begin{cases}0, & x \in I \\ {\left[x^{2}\right]} & \text { otherwise }\end{cases}$

Short Answer

Expert verified
Question: Give the expressions for the compositions of functions g(f(x)) and f(g(x)), when f(x) is the greatest integer function, and g(x) is a piecewise function given by \(g(x)= \begin{cases}0, & x \in I \\\ x^{2}, & \text { otherwise }\end{cases}\) where I is an unspecified interval. Answer: g(f(x)) = 0 for all x in R f(g(x)) = \( \begin{cases}0, & x \in I \\\ {\left[x^{2}\right]} & \text { otherwise }\end{cases}\)

Step by step solution

01

Understanding the given functions

We're given two functions 鈥� f(x) and g(x): \(f(x)=\text{[}x\text{]}\) is the greatest integer function, which means the greatest integer less than or equal to x. \(g(x)= \begin{cases}0, & x \in I \\\ x^{2}, & \text { otherwise }\end{cases}\) is a piecewise function with two cases: when x belongs to a specific interval I, g(x) = 0; when x does not belong to that interval I, g(x) = x虏. We are not given the specific interval 'I'. It could be any interval within the domain.
02

Finding g(f(x))

To find \(g(f(x))\), we will substitute \(f(x)=[x]\) into \(g(x)\): \(g(f(x))= \begin{cases}0, & f(x) \in I \\\ \left(f(x)\right)^{2}, & \text{otherwise }\end{cases}\) Since the exercise states that \(g(f(x)) = 0\) for all \(x\in R\), it implies that \(f(x) \in I\) for all x. And the expression for \(g(f(x))\) will be: \(g(f(x))=0 \ \text{for all} \ x \in R\)
03

Finding f(g(x))

To find \(f(g(x))\), we will substitute \(g(x)\) into \(f(x)\): \(f(g(x))=[g(x)]\) Now, we need to consider the two cases for \(g(x)\): Case 1: When \(x \in I\), \(g(x)=0\), So, \(f(g(x))=[0]=0\) Case 2: When \(x \notin I\), \(g(x)=x^2\) So, \(f(g(x))=[x^2]\) Combine both cases to write the expression for \(f(g(x))\): \(f(g(x))= \begin{cases}0, & x \in I \\\ {\left[x^{2}\right]} & \text { otherwise }\end{cases}\)

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Most popular questions from this chapter

$\lim _{x \rightarrow 0} \frac{\sqrt{4-2 x+x^{2}}-\sqrt{4+2 x+x^{2}}}{\sqrt{2+x}-\sqrt{2-x}}$ Rationalize, $\lim _{x \rightarrow 0} \frac{(-4 x)(\sqrt{2+x}+\sqrt{2-x})}{(2 x)\left(\sqrt{4-2 x+x^{2}}+\sqrt{4+2 x+x^{2}}\right)}$ \(\Rightarrow-\sqrt{2}\) Hence, \(\mathrm{C}\) is correct

$\mathrm{f}(\mathrm{x})=\lim _{\mathrm{n} \rightarrow \infty} \frac{(2 \sin \mathrm{x})^{2 \mathrm{n}}}{3^{\mathrm{n}}-(2 \cos \mathrm{x})^{2 \mathrm{n}}}\( \)\Rightarrow \lim _{n \rightarrow \infty} \frac{(2 \sin x)^{2 n}}{1-\left(\frac{2}{\sqrt{3}} \cos x\right)^{2 n}}$ \(\mathrm{f}(\mathrm{x})\) is discontinuous whenever \(2 \sin x=\pm 1 \& \frac{2}{\sqrt{3}} \cos x=\pm 1\) \(\Rightarrow \mathrm{x}=\mathrm{n} \pi \pm \frac{\pi}{6}\)

Correction $\rightarrow f(x)=\frac{1-\cos x(\cos 2 x)^{1 / 2} \cos (3 x)^{1 / 3}}{x^{2}}$ Sol using L-Hospital's Rule $\begin{aligned} & \lim _{x \rightarrow 0}(\sin x)(\cos 2 x)^{1 / 2}(\cos 3 x)^{1 / 3}+\cos x(\cos 3 x)^{1 / 3} \\ \lim _{x \rightarrow 0} f(x)=& \frac{\sin 2 x}{\sqrt{\cos 2 x}+\cos x(\cos 2 x)^{1 / 3} \frac{\sin 3 x}{(\cos 3 x)^{2 / 3}}}{2} \\\=& \lim _{x \rightarrow 0}\left[\frac{1}{2}+1+\frac{3}{2}\right]=3 \end{aligned}$

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

\(\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\tan k x}{x}=k\) $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}\left(3+2 k^{2}\right)=2 k^{2}$ For continuity, \(\mathrm{k}=2 \mathrm{k}^{2}\) \(\Rightarrow \mathrm{K}=0\) or \(\frac{1}{2}\) Hence, \(\mathrm{C}\) is correct.
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