/*! 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 84 \(f^{\prime}(x)=\frac{-2}{x^{2}}... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 84

\(f^{\prime}(x)=\frac{-2}{x^{2}}\left(\frac{\sqrt{x^{2}+4}+2}{\sqrt{x^{2}+4}}+2\right)<0\) Hence, A is correct

Short Answer

Expert verified
Answer: Yes, the derivative of the function is always less than zero.

Step by step solution

01

Rewrite the expression for better understanding

Let's rewrite the given expression to make it easier to work with: \(f^{\prime}(x) = \frac{-2}{x^{2}} \left(\frac{\sqrt{x^{2} + 4} + 2}{\sqrt{x^{2} + 4}} + 2\right) < 0\)
02

Distribute the \(\frac{-2}{x^{2}}\) term across the parentheses

Distribute the \(\frac{-2}{x^{2}}\) term to both terms inside the parentheses: \(f^{\prime}(x) = \frac{-2 (\sqrt{x^{2} + 4} + 2)}{x^{2}\sqrt{x^{2} + 4}} + \frac{-4}{x^{2}} < 0\)
03

Combine the terms with a common denominator

Combine the terms in the expression by finding a common denominator, which is \(x^{2}\sqrt{x^{2} + 4}\): \(f^{\prime}(x) = \frac{-2 (\sqrt{x^{2} + 4} + 2) - 4\sqrt{x^{2} + 4}}{x^{2}\sqrt{x^{2} + 4}} < 0\)
04

Simplify the numerator

We can simplify the numerator further by distributing the -2 and combining like terms: \(f^{\prime}(x) = \frac{-2\sqrt{x^{2} + 4} - 4 - 4\sqrt{x^{2} + 4}}{x^{2}\sqrt{x^{2} + 4}} < 0\) \(f^{\prime}(x) = \frac{-6\sqrt{x^{2} + 4} - 4}{x^{2}\sqrt{x^{2} + 4}} < 0\)
05

Determine the sign of the expression

Now let's determine the sign of the expression for different values of x. Since the denominator is always positive, we only need to consider the numerator: 1. For x > 0: The term inside the square root \(\sqrt{x^2+4}\) is always positive, so the numerator is negative. 2. For x < 0: The same reasoning holds, as the term inside the square root is always positive, so the numerator is negative. Therefore, the expression is negative for all x, and we can conclude that \(f^{\prime}(x) < 0\). Hence, the correct answer is A.

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Most popular questions from this chapter

A) \(\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}+\frac{4}{\mathrm{x}^{2}}+7\) \(f^{\prime}(x)=3 x^{2}-\frac{8}{x^{3}}=\frac{3 x^{5}-8}{x^{3}}\) \(\mathrm{f}(\mathrm{x})\) is \(\uparrow\) in \((-\infty, \infty)\) B) \(g(x)=\sqrt{t}+\sqrt{1+t}-4\) \(g^{\prime}(x)=\frac{1}{2 \sqrt{t}}+\frac{1}{2 \sqrt{1+t}}>0\) \(\mathrm{g}(\mathrm{x})\) is increasing in \((0, \infty)\) C) \(\mathrm{r}(\theta)=\theta+\sin ^{2}\left(\frac{\theta}{3}\right)-8\) \(r^{\prime}(\theta)=1+\frac{1}{3} \sin \left(\frac{2 \theta}{3}\right)>0\) \(\mathrm{r}(\theta)\) is an increasing function D) \(r(\theta)=\tan \theta-\cot \theta-\theta\) \(r^{\prime}(\theta)=\sec ^{2} \theta+\operatorname{cosec}^{2} \theta-1\) \(=\tan ^{2} \theta+\operatorname{cosec}^{2} \theta\) \(\mathrm{r}(\theta)\) is an increasing function Hence, $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$ is correct

$g^{\prime}(\theta)=\left(f\left(\sin ^{2} \theta\right)-f\left(\cos ^{2} \theta\right)\right) \sin 2 \theta$ In $\left(-\frac{\pi}{4}, \theta\right), \sin 2 \theta<0 \& \sin ^{2} \theta<\cos ^{2} \theta$ \& \(f\left(\sin ^{2} \theta\right)0\) Hence, D is correct

\(f^{\prime \prime}(x)>0 \Rightarrow f^{\prime}(x)\) is increasing \(g(x)=f\left(\tan ^{2} x-2 \tan x+4\right)\) $g^{\prime}(x)=f^{\prime}\left(\tan ^{2} x-2 \tan x+4\right)\left(2 \tan x \sec ^{2} x-2 \sec ^{2} x\right)$ \(=f^{\prime}\left((\tan x-1)^{2}+3\right) 2 \sec ^{2} x(\tan x-1)\) As \(\left.f^{\prime}(\tan x-1)^{2}+3\right)\) is \(+v e\) \(\Rightarrow \mathrm{g}^{\prime}(\mathrm{x})>0\) if $\mathrm{x} \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$ Hence, D is correct

$f^{\prime}(x)= \begin{cases}\frac{1}{2 \sqrt{x}}, & x \geq 1 \\ 3 x^{2}, & 0 \leq x \leq 1 \\ x^{2}-4, & x<0\end{cases}$ \(\mathrm{f}^{\prime}(\mathrm{x})\) changes its sign at $\mathrm{x}=0 \& \mathrm{c}-2$ Hence, \(\mathrm{C}\) is correct

\(f(x)=x^{3}+2 x^{2}+5 x+2 \cos x\) \(f^{\prime}(x)=3 x^{2}+4 x+5-2 \sin x\) Min value of \(3 \mathrm{x}^{2}+4 \mathrm{x}+5\) is \(\frac{22}{3}\) \(\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})>0\) \(f(0)=2 \& f(\infty)=\infty, \quad f(-\infty)-\infty\) There is no root in \([0,2 \pi]\) Hence, D is correct
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