/*! 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 62 \(f^{\prime}(x)=g(x)(x-a)^{2}\) ... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 62

\(f^{\prime}(x)=g(x)(x-a)^{2}\) \(f(x)\) is increasing if \(g(a)>0\) \& \(\mathrm{f}(\mathrm{x})\) is decreasing if \(\mathrm{g}(\mathrm{a})<0\) Hence, A, D is correct

Short Answer

Expert verified
A. \(f(x)\) is increasing if \(g(a)>0\) B. \(f(x)\) is increasing if \(g(a)<0\) C. \(f(x)\) is decreasing if \(g(a)>0\) D. \(f(x)\) is decreasing if \(g(a)<0\) Answer: A and D

Step by step solution

01

Utilize the First Derivative Test

To determine if \(f(x)\) is increasing or decreasing, we'll rely on the First Derivative Test. This test states that: 1. If \(f'(x) > 0\) for all \(x\) in an interval, then \(f(x)\) is increasing in that interval. 2. If \(f'(x) < 0\) for all \(x\) in an interval, then \(f(x)\) is decreasing in that interval.
02

Analyze the given condition for an increasing function and its derivative

We are given that \(f(x)\) is increasing if \(g(a) > 0\). In order for this to be true, \(f'(x)\) must be positive for the given value of \(x\). We have \(f'(x) = g(x)(x-a)^2\). Since the square of any real number \((x-a)^2\) is always non-negative, the whole expression for \(f'(x)\) would be positive only if \(g(x) > 0\).
03

Analyze the given condition for a decreasing function and its derivative

Similarly, we are given that \(f(x)\) is decreasing if \(g(a) < 0\). In order to fulfill this condition, we must have \(f'(x) < 0\) for the given value of \(x\). Again, since \((x-a)^2\) is always non-negative, the only way for \(f'(x)\) to be negative in this case is if \(g(x) < 0\).
04

Conclusion

Based on the first derivative test and the analysis of the given conditions, we can conclude that the statements A and D are correct: - \(f(x)\) is increasing if \(g(a)>0\) - \(f(x)\) is decreasing if \(g(a)<0\)

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Most popular questions from this chapter

\(f(x)=\sin ^{2} x-3 \cos ^{2} x+2 a x-4\) \(f^{\prime}(x)=4(\sin 2 x)+2 a \geq 0\) \(\Rightarrow \mathrm{a} \in[2, \infty)\) Hence, \(\mathrm{C}\) is correct

$\left.\mathrm{f}^{\prime}(\mathrm{x})=\log _{1 / 3} \log _{3}(\sin (\mathrm{x})+\mathrm{a})\right)<0$ \(\left.\log _{3}(\sin (x)+a)\right)>1\) \(\sin (x)+a>3\) \(a>3-\sin x\) \(\mathrm{a}>4\) Hence B is correct

\(x=t^{2} \quad \& y=3 t+t^{3}\) \(\frac{d x}{d t}=2 t \quad \& \quad \frac{d y}{d t}=3+3 t^{2}\) \(\frac{d y}{d x}=\frac{3}{2} \frac{\left(1+t^{2}\right)}{t}\) $\frac{d^{2} y}{d x^{2}}=\frac{3}{2} \frac{t(2 t)-\left(1+t^{2}\right)}{t^{2}} \times \frac{1}{2 t}$ \(=\frac{3}{4} \frac{\left(t^{2}-1\right)}{t^{3}}\) $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}} \rightarrow \frac{-}{ }_{-1}^{+} \underset{0}{1} \stackrel{+}{ }_{1}^{+}$ 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

$\mathrm{f}(\mathrm{x})=(4 \mathrm{a}-3)(\mathrm{x}+5)+2(\mathrm{a}-7) \cot \frac{\mathrm{x}}{2} \sin ^{2} \frac{\mathrm{x}}{2}$ \(f(x)=(4 a-3)(x+5)+(a-7) \sin x\) \(f^{\prime}(x)=4 a-3+(a-7) \cos x\) \(\Rightarrow 4 \mathrm{a}-3>-(\mathrm{a}-7)\) or \(4 \mathrm{a}-3<\mathrm{a}-7\) \(5 a>10\) or \(3 a<-4\) \(a>2\) or \(a<-4 / 3\) \(\mathrm{a} \in\left(-\infty,-\frac{4}{3}\right) \cup(2, \infty)\) Hence, D is correct
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