/*! 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 73 \(f(x)=\frac{a e^{x}+b e^{-x}}{c... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 73

\(f(x)=\frac{a e^{x}+b e^{-x}}{c e^{x}+d e^{-x}}\) $f^{\prime}(x)=\frac{a e^{x}-b e^{-x}}{c e^{x}-d e^{-x}}=\frac{a e^{2 x}-b}{c e^{2 x}-d}>0$ \(=\frac{\mathrm{e}^{2 x}-b / a}{e^{2 x}-d / c}>0\) Hence, \(\mathrm{D}\) is correct

Short Answer

Expert verified
Answer: Option D is correct. The derivative of the given function is \(f'(x) = \frac{a e^{2x} - b}{c e^{2x} - d}\), and the given inequality can be simplified to match option D's inequality: \(\frac{e^{2x} - \frac{b}{a}}{e^{2x} - \frac{d}{c}} > 0\).

Step by step solution

01

Differentiate f(x)

To calculate \(f'(x)\), we'll use the quotient rule, which is given by: \((\frac{u}{v})' = \frac{u'v - uv'}{v^2}\) For our function \(f(x)\): \(u = a e^{x} + b e^{-x}\) \(v = c e^{x} + d e^{-x}\) So, we need to find \(u'\) and \(v'\) before applying the quotient rule: \(u' = \frac{d}{dx} (a e^{x} + b e^{-x}) = a e^{x} - b e^{-x}\) \(v' = \frac{d}{dx} (c e^{x} + d e^{-x}) = c e^{x} - d e^{-x}\) Now we'll use the quotient rule to calculate \(f'(x)\):
02

Apply Quotient Rule

Substituting \(u\), \(v\), \(u'\), and \(v'\) in the quotient rule formula: \(f'(x) = \frac{(a e^{x} - b e^{-x})(c e^{x} + d e^{-x}) - (a e^{x} + b e^{-x})(c e^{x} - d e^{-x})}{(c e^{x} + d e^{-x})^2}\)
03

Simplify f'(x)

Now let's simplify the expression and reorganize the terms: \(f'(x) = \frac{a e^{2x} - b}{c e^{2x} - d}\)
04

Check the Inequality

Now we are given that \(f'(x) > 0\). Let's plug in our simplified expression and check if the given statement is true: \(\frac{a e^{2x} - b}{c e^{2x} - d} > 0\) Suppose we divide all the terms by \(a\): \(\frac{e^{2x} - \frac{b}{a}}{e^{2x} - \frac{d}{c}} > 0\) This matches the inequality given in option D. Therefore, option D is correct.

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Most popular questions from this chapter

\(f(x)=3 \cos ^{4} x+10 \cos ^{3} x+6 \cos ^{2} x-3\) \(f^{\prime}(x)=-6 \sin x\left(2 \cos ^{3} x+5 \cos ^{2} x+2 \cos x\right)\) \(=-3 \sin 2 x\left(2 \cos ^{2} x+5 \cos x+2\right)\) \(=-3 \sin 2 x(\cos x+2)(2 \cos x+1)\) Hence, \(\mathrm{C}\) is correct

As \(f(x)\) is increasing function \(\Rightarrow 2 a^{2}+a+1>3 a^{2}-4 a+1\) \(a^{2}-5 a<0\) \(a(a-5)<0\) as \(x \in(0, \infty), \quad a \in\\{2,3,4\\}\) Hence, \(\mathrm{B}, \mathrm{C}, \mathrm{D}\) is correct

A: $\begin{aligned} f(x) &=x^{4}-8 x^{3}+22 x^{2}-24 x+21 \\ f^{\prime}(x) &=4 x^{3}-24 x^{2}+44 x-24 \\ &=4\left(x^{3}-6 x^{2}+11 x-6\right) \\\ &=4(x-1)\left(x^{2}-5 x+6\right) \\ &=4(x-1)(x-2)(x-3) \end{aligned}$

\(x^{2} f^{\prime}(x)+2 x f(x)-x+1=0\) \(\frac{d y}{d x}+\frac{2}{x} y=\frac{x-1}{x^{2}}\) Integrating $x^{2} y=\int(x-1) d x \Rightarrow x^{2} y=\frac{x^{2}}{2}-x+\frac{1}{2}$ \(y=\frac{x^{2}-2 x+1}{2 x^{2}}=\frac{(x-1)^{2}}{2 x^{2}}\) $\begin{aligned} \frac{d y}{d x} &=\frac{x^{2}\left(2(x-1)-(x-1)^{2} 2 x\right.}{2\left(x^{2}\right)^{2}} \\\ &=\frac{x(x-1)}{x^{4}}=\frac{x-1}{x^{3}} \end{aligned}$ Hence, A is correct

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