/*! 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 39 \(f(x)=a x^{3}+b x^{2}+c x+d\) ... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 39

\(f(x)=a x^{3}+b x^{2}+c x+d\) \(f^{\prime}(x)=3 a x^{2}+2 b x+c>0\) \(\Rightarrow 4 b^{2}-12 a c<0 \& a>0\) \(\Rightarrow \mathrm{b}^{2}-3 \mathrm{ac}<0\) \(b^{2}<3 a c \& a>0,3 b^{2}\left(\sqrt{3} a x+\frac{4 b}{\sqrt{3}}\right)^{2}>0\) Hence, B is correct

Short Answer

Expert verified
Answer: The relation between h(x) and g(x) is that h(x) is an antiderivative of g(x), which means h'(x) = g(x). Additionally, g(x) satisfies the condition \(g(x) > \left(\sqrt{3}ax + \frac{4b}{\sqrt{3}}\right)^2 > 0\).

Step by step solution

01

Analyze Conditions for \(f'(x)\)

We are given that \(f'(x)>0\), which indicates that the function is strictly increasing. From the provided conditions, we have: 1. \(4b^2 - 12ac < 0 \Rightarrow b^2 < 3ac\) 2. \(a > 0\) 3. \(3b^2 < c^2\) These conditions help us understand the constraints on the coefficients \(a\), \(b\), and \(c\).
02

Determine \(g(x)\)

Using the conditions found above in the exercise, we have the following: \(g(x) = a(3ax^2 + 2bx + c) + b(6ax + 2b) + c^2\) Expanding this, we get: \(g(x) = 3a^2x^2 + 2abx + \frac{b^2}{3} + 6abx + 2b^2 + 3b^2\) Which simplifies to: \(g(x) > \left(\sqrt{3}ax + \frac{4b}{\sqrt{3}}\right)^2 > 0\)
03

Integrate \(g(x)\) to find \(h(x)\)

Now that we have determined the function \(g(x)\), we can integrate to find \(h(x)\): \(h(x) = \int_{a}^{x} g(t) dt\) Since \(g(x) > \left(\sqrt{3}ax + \frac{4b}{\sqrt{3}}\right)^2 > 0\), the integral will be strictly positive, and the resulting function \(h(x)\) will satisfy the condition \(h'(x)=g(x)\). Hence, the correct option is B, which states that \(h'(x) = g(x)\) and \(g(x) > \left(\sqrt{3}ax + \frac{4b}{\sqrt{3}}\right)^2 > 0\).

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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

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

\(x^{2} f^{\prime}(x)+2 x f(x)-x+1=0\) \(\frac{d y}{d x}+\frac{2}{x} y=\frac{x-1}{x^{2}}\) 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

If \(f^{\prime \prime}(x)>0 \Rightarrow f^{\prime}(x)\) is increasing function \(g(x)=f(2-x)+f(4+x)\) \(g^{\prime}(x)=f^{\prime}(4+x)-f^{\prime}(2-x)>0\) \(\Rightarrow \quad 4+x>2-x\) \(2 x>-2\) \(x>-1\) Hence, \(\mathrm{C}\) is correct.

\(f(x)=g(x)\) \(\sin 3 x+\cos x=\cos 3 x+\sin x\) \(\sin 3 x-\sin x=\cos 3 x-\cos x\) \(\cos 2 x \sin x=-\sin 2 x \sin x\) \(\Rightarrow \sin x(\cos 2 x+\sin 2 x)=0\) \(\Rightarrow \sin x \sin \left(\frac{\pi}{4}+2 x\right)=0\) \(\Rightarrow\) either \(\sin x=0\) or \(\sin \left(\frac{\pi}{4}+2 x\right)=0\) \(x=0, \pi, \frac{3 \pi}{8}, \frac{7 \pi}{8}\) Hence, \(\mathrm{C}\) is correct
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