/*! 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 28 \(f(x)+2 f(-x)=-\sin x\) Put \... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 28

\(f(x)+2 f(-x)=-\sin x\) Put \(x=-x\) \(f(-x)+2 f(x)=-\sin x\) Solve, (I) \& (II), we get \(f(x)=-\sin x\) \(f^{\prime}(x)=-\cos x\) \(\mathrm{f}^{\prime}\left(\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}}\)

Short Answer

Expert verified
Answer: f'(蟺/4) = -1/鈭�2

Step by step solution

01

Solve the given functional equation

The given functional equation is: \(f(x)+2 f(-x)=-\sin x\) Let's call it Equation (I). Now, we are also given that if we substitute \(x=-x\), we get another equation: \(f(-x)+2 f(x)=-\sin x\) Let's call it Equation (II). Now, we need to solve Equation (I) and Equation (II) to find the function f(x).
02

Solve Equation (I) and Equation (II) simultaneously

To find the function f(x), we will solve Equations (I) and (II) simultaneously. Adding Equation (I) and Equation (II): \(( f(x)+2 f(-x)) + (f(-x)+2 f(x)) = -\sin x + (-\sin x)\) Simplifying the equation: \(3f(x) + 3f(-x) = -2\sin x\) Now, divide both sides by 3: \(f(x) + f(-x) = -\frac{2}{3}\sin x\) From Equation (I), we already have: \(f(x) + 2f(-x) = -\sin x\) Now subtract the previous equation from this one: (2f(-x) - f(-x)) = (-\sin x - (-\frac{2}{3}\sin x))$ This simplifies to: \(f(-x) = -\frac{1}{3}\sin x\) Now, substituting this expression for f(-x) back into Equation (I): \(f(x) + 2(-\frac{1}{3}\sin x) = -\sin x\) Simplifying, we get: \(f(x) = -\sin x\)
03

Differentiate f(x)

Now that we have found f(x), we will differentiate it with respect to x to find f'(x). \(f(x) = -\sin x\) Using the chain rule, we get: \(f'(x) = -\cos x\)
04

Find f'(蟺/4)

And finally, we want to find the value of f'(x) when x = 蟺/4. So, we substitute 蟺/4 into f'(x): \(f'\left(\frac{\pi}{4}\right) = -\cos\left(\frac{\pi}{4}\right)\) Since \(\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}\): \(f'\left(\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}}\)

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Most popular questions from this chapter

\(y=\log _{e^{t}}(x-2)^{2}\) \(y=\frac{2 \log (x-2)}{\log x}\) $y^{\prime}=\frac{2(\log x) \times \frac{1}{x-2}-2 \log (x-2) \times \frac{1}{x}}{(\log x)^{2}}$ \(y^{\prime}=\frac{2}{\log 3}\) at \(x=3\)

\(x=t \cos t, y=t+\sin t\) \(\frac{d x}{d t}=\cos t-t \sin t, \frac{d y}{d t}=1+\cos t\) \(\frac{d y}{d x}=\frac{1+\cos t}{\cos t-t \sin t}\) \(\frac{d x}{d y}=\frac{\cos t-t \sin t}{1+\cos t}\) $\frac{d^{2} x}{d y^{2}}=\frac{(1+\cos t)(-\sin t-\sin t-t \cos t)+(\cos t-t \sin t) \sin t}{(1+\cos t)^{3}}$ \(=-2-\pi / 2\) \(=-\frac{(\pi+4)}{2}\)

Assume that \(f\) is differentiable for all \(x\). The sign of \(f^{\prime}\) is as follows: \(\mathrm{f}^{\prime}(\mathrm{x})>0\) on \((-\infty,-4)\) \(\mathrm{f}^{\prime}(\mathrm{x})<0\) on \((-4,6)\) \(\mathrm{f}^{\prime}(\mathrm{x})>0\) on \((6, \infty)\) Let \(g(x)=f(10-2 x)\). The value of \(g^{\prime}(L)\) is (A) positive (B) negative (C) zero (D) the function \(g\) is not differentiable at \(x=5\)

$\begin{aligned} &\mathrm{y}=(1-\mathrm{x})^{-\alpha} \mathrm{e}^{-\alpha \mathrm{x}}\\\ &\text { Taking log on both sides, }\\\ &\ln y=-\alpha \ln (1-x)-\alpha x\\\ &\Rightarrow \frac{1}{y} y^{\prime}=\frac{+\alpha}{1-x}-\alpha\\\ &\Rightarrow(1-x) y^{\prime}=\alpha y-\alpha y(1-x)\\\ &\Rightarrow(1-x) y^{\prime}=\alpha x y\\\ &\Rightarrow(1-x) y^{\prime \prime}-y^{\prime}=\alpha x y^{\prime}+\alpha y\\\ &\Rightarrow(1-x) y^{\prime \prime}-(1+\alpha x) y^{\prime}-\alpha y=0 \end{aligned}$

Let the function \(f\) satisfy the relation, \(f\left(x+y^{3}\right)=f(x)+f\left(y^{3}\right)\), $\forall \mathrm{x}, \mathrm{y} \in \mathrm{R}\( and be differentiable for all \)\mathrm{x}$. Assertion \((\mathbf{A}):\) If \(f^{\prime}(2)=a\), then \(f^{\prime}(-2)=a\). Reason \((\mathbf{R}): \mathrm{f}(\mathrm{x})\) is an odd function
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