/*! 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 13 \(f(x)=\frac{x^{3}}{4}-a \sin \p... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 13

\(f(x)=\frac{x^{3}}{4}-a \sin \pi x+3, \quad f(4)=19, \quad f(-4)=-13\) As \(\mathrm{f}(\mathrm{x})\) is continuous \(\&-13<\frac{1999}{199}<19\). Using IVT, \(f(x)\) is \(\frac{1999}{199}\) for some \(x \in[-4,4]\)

Short Answer

Expert verified
Additionally, verify that there exists a value of \(x\) in the interval \([-4, 4]\) such that \(f(x) = \frac{1999}{199}\). Answer: The value of 'a' is 0, and there exists at least one value \(x\) in the interval \([-4, 4]\) such that \(f(x) = \frac{1999}{199}\).

Step by step solution

01

Use the given values to form equations to solve for a

We are given that \(f(4) = 19\) and \(f(-4) = -13\). Let's plug the values of \(x\) into the function and equate them to the given results. For \(x=4\), we get the equation: \(f(4) = \frac{4^3}{4} - a\sin(\pi(4)) + 3 = 19\) For \(x=-4\), we get the equation: \(f(-4) = \frac{(-4)^3}{4} - a\sin(\pi(-4)) + 3 = -13\)
02

Solve for 'a'

Solve the two equations obtained in Step 1 to find the value of 'a'. Equation 1: \(\frac{4^3}{4} - a\sin(\pi(4)) + 3 = 19\) Since \(\sin(\pi x)\) is 0 for any integer value of \(x\), the term involving \(a\) becomes 0 Solving for 'a', we get \(a = \frac{4^3}{4} + 3 - 19\) Equation 2: \(\frac{(-4)^3}{4} - a\sin(\pi(-4)) + 3 = -13\) Similarly, the term involving \(a\) becomes 0 in this equation as well Solving for 'a', we get \(a = \frac{(-4)^3}{4} + 3 + 13\) By comparing both equations, we can see that they have the same value for 'a'.
03

Compute the value of 'a'

Now, let's compute the value of 'a' using any of the above equations. \(a = \frac{4^3}{4} + 3 - 19 = 16 + 3 - 19 = 0\) So, the value of 'a' is 0.
04

Plug the value of 'a' into the function

Now that we have found the value of 'a', we can plug it into the function \(f(x)\). \(f(x) = \frac{x^3}{4} - 0\sin(\pi x) + 3 = \frac{x^3}{4} + 3\)
05

Verify using IVT

We need to verify that there exists a value of \(x\) in the interval \([-4, 4]\) such that \(f(x) = \frac{1999}{199}\). Since \(f(x)\) is a continuous function and \(a=0\), the function reduces to \(f(x) = \frac{x^3}{4} + 3\). Let's find \(f(-4)\) and \(f(4)\). \(f(-4) = \frac{(-4)^3}{4} + 3 = -16 + 3 = -13\) \(f(4) = \frac{4^3}{4} + 3 = 16 + 3 = 19\) Since \(f(-4)=-13\) and \(f(4)=19\), and \(-13 < \frac{1999}{199} < 19\), by the Intermediate Value Theorem, there exists at least one value of \(x\) in the interval \([-4,4]\) such that \(f(x) = \frac{1999}{199}\).

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Most popular questions from this chapter

$\mathrm{f}(\mathrm{x})= \begin{cases}\sin \mathrm{x}, & \mathrm{x} \in \text { Rational } \\ 1-2 \cos \mathrm{x}, & \mathrm{x} \in \text { Irrational }\end{cases}$ For continuity, \(\sin x=1-2 \cos x\) \(\sin x+2 \cos x=1\) \(\frac{1}{\sqrt{5}} \sin x+\frac{2}{\sqrt{5}} \cos x=\frac{1}{\sqrt{5}}\) \(\sin (x+\alpha)=\frac{1}{\sqrt{5}}\), where \(\cos \alpha=\frac{1}{\sqrt{5}}\) There are infinite values of ' \(x\) ' which satisfies the eqn.

\(\mathrm{f}(\mathrm{x})=(\mathrm{x} \bmod 2)^{2}+(\mathrm{x} \bmod 4)\) Points of discontinuity are \(\mathrm{x}=2,4,6,8\) Hence, \(\mathrm{C}\) is correct.

As $g(x)= \begin{cases}(a-x) \sec \frac{\pi x}{2 a}, & xa\end{cases}$ $\begin{aligned} \lim _{x \rightarrow a} g(x) &=\lim _{x \rightarrow a} \frac{(a-x)}{\cos \left(\frac{\pi x}{2 a}\right)} \\ &=\lim _{b \rightarrow 0} \frac{h}{\sin \left(\frac{\pi}{2} \frac{h}{a}\right)} \Rightarrow \frac{2 a}{\pi} \\ \lim _{x \rightarrow a^{\prime}} g(x)=\lim _{h \rightarrow 0}\left(-\cot \left(\frac{\pi}{2}+\frac{\pi}{2 a} h\right) \operatorname{cosec}(h)\right) \\ &=\lim _{h \rightarrow 0}+\frac{\tan \left(\frac{\pi}{2 a} h\right)}{\sin h} \times \frac{h}{\frac{\pi}{2 a} h} \times \frac{\pi}{2 a}=\frac{\pi}{2 a} \end{aligned}$ Hence, D is correct.

\(\mathrm{f}(\mathrm{x})=\) highest power of \(\left(\mathrm{u}^{\mathrm{x}^{2}}+\mathrm{u}^{2}+2 \mathrm{u}+3\right)\) $\lim _{\mathrm{x} \rightarrow \sqrt{2}} \mathrm{f}(\mathrm{x})=2 \quad \mathrm{f}(\sqrt{2})=2$ \(\lim _{x \rightarrow \sqrt{2}} \mathrm{f}(\mathrm{x})=2\)

If \(x \geq 0\) \(x^{2}-x+1-2=0\) \(x^{2}-x-1=0\) \(x=1 \pm \sqrt{5} \Rightarrow x=\frac{1+\sqrt{5}}{2}\) If \(x<0\) \(x^{2}-x-2=0\) \((x-2)(x+1)=0\) \(\mathrm{x}=2\) or \(-1\) so, \(x=-1\) No options matches.
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