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\(f(x)+2 f(-x)=-\sin x\) 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'(π/4) = -1/√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'(π/4)

And finally, we want to find the value of f'(x) when x = π/4. So, we substitute π/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

Column-I (A) If \(|g(x)-g(y)| \leq|x-y|^{99},(n \in N)\) then \(g^{\prime}(x)\) equals (B) Differential coefficient of $\sin ^{-1} \frac{2 \mathrm{x}}{1+\mathrm{x}^{2}}\( with respect to \)\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}\( in the domain of \)f(x)=\frac{1}{\sqrt{1-x^{2}}}$ is (C) If \(y=\cot ^{-1}(\tan x)+\tan ^{-1}(\cot x), \pi / 2

\(f(x)=\frac{x}{1+e^{1 / x}}\) \(f(x)\left(1+e^{1 / x}\right)-x=0\) \(f^{\prime}(x)\left(1+e^{1 / x}\right)-\frac{f(x) e^{l / x}}{x^{2}}-1=0\) \(x^{2} f^{\prime}(x) \frac{(x)}{f(x)}-f(x) e^{1 / x}-x^{2}=0\)

If for some differentiable function \(\mathrm{f}, \mathrm{f}(\alpha)=0\) and \(\mathrm{f}^{\prime}(\alpha)=0\). Assertion (A) : The sign of \(\mathrm{f}(\mathrm{x})\) does not change in the neighbourhood of \(\mathrm{x}=\alpha\). Reason \((\mathbf{R}): \alpha\) is repeated root of \(\mathrm{f}(\mathrm{x})=0\)

The value of $\frac{\mathrm{f}(\mathrm{t})}{\mathrm{f}^{\prime}(\mathrm{t})} \cdot \frac{\mathrm{f}^{\prime \prime}(-\mathrm{t})}{\mathrm{f}^{\prime}(-\mathrm{t})}+\frac{\mathrm{f}(-\mathrm{t})}{\mathrm{f}^{\prime}(-\mathrm{t})} \cdot \frac{\mathrm{f}^{\prime \prime}(\mathrm{t})}{\mathrm{f}^{\prime}(\mathrm{t})}$ \(\forall \mathrm{t} \in \mathrm{R}\), is equal to (A) \(-2\) (B) 2 (C) \(-4\) (D) 4

As $\mathrm{f}(\mathrm{x}), \mathrm{f}^{\prime}(\mathrm{x}), \mathrm{f}^{\prime \prime}(\mathrm{x})\( are all \)+\mathrm{ve} \forall \mathrm{x} \in[0,7]$ \(\Rightarrow \mathrm{f}(\mathrm{x})\) is increasing function and concave up. and so is \(\mathrm{f}^{-1}(\mathrm{x})\). $\Rightarrow f^{-1}(5)+4 f^{-1}\left(\frac{2 g}{5}\right)\( is always \)+v e$.

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