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Chapter 10: Problem 801

If \(f\) is an even function and \(f^{\prime}(x)\) is define than \(f^{\prime}(x)+f^{\prime}(-x)\) (a) 0 (b) \(<0\) \((\mathrm{c}) \neq 0\) \((\mathrm{d})>0\)

Short Answer

Expert verified
(a) 0

Step by step solution

01

Apply the definition of even function

Given that \(f\) is an even function, we have the equation \(f(-x) = f(x)\). Differentiate both sides with respect to \(x\).
02

Differentiate the equation

Derivate both sides: \[\frac{d}{dx}f(-x) = \frac{d}{dx}f(x)\] Apply the chain rule on the left-hand side: \[-f'(-x) = f'(x)\]
03

Find the value of f'(x) + f'(-x)

We are asked to find the value of \(f'(x) + f'(-x)\). To do this, we can simply add the right-hand side and left-hand side of the equation we got in step 2: \[-f'(-x) + f'(-x) = f'(x) + f'(x)\] Now, observe that we can simplify this equation: \[0 = 2f'(x)\]
04

Determine the choice of value

Since we have \(0 = 2f'(x)\), we can divide both sides by 2 to get: \[0 = f'(x)\] Therefore, the value of \(f'(x) + f'(-x)\) is 0. So, the correct answer is: (a) 0

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Most popular questions from this chapter

If \(y=\cos ^{-1} \mid\left[\left(3 x+4 \sqrt{ \left.\left(1-x^{2}\right)\right) / 5}\right] \mid\right.\) then \((\operatorname{dy} / d x)=\) (a) \(\left.\left[\\{-1\\} /\left\\{\sqrt{(} 1-\mathrm{x}^{2}\right)\right\\}\right]\) (b) \(\left[2 /\left\\{\sqrt{\left. \left.\left(1-\mathrm{x}^{2}\right)\right\\}\right]}\right.\right.\) (c) \(\left[5 /\left\\{3 \sqrt{\left. \left.\left(1-x^{2}\right)\right\\}\right]}\right.\right.\) (d) \(\left[\\{-3\\} /\left\\{5 \sqrt{\left. \left.\left(1-x^{2}\right)\right\\}\right]}\right.\right.\)

If \(x=\left[\left(e^{2 y}+1\right) /\left(e^{2 y}-1\right)\right]\) then \((d y / d x)=\) (a) \(\left[1 /\left(1+\mathrm{x}^{2}\right)\right]\) (b) \(\left[1 /\left(\mathrm{x}^{2}-1\right)\right]\) (c) \(\left[1 /\left(1-x^{2}\right)\right]\) (d) \(\left[1 /\left(x^{2}-1\right)\right]\)

If \(\mathrm{y}=\log \sqrt{\\{}(1-\cos \mathrm{ax}) /(1+\cos a \mathrm{x})\\}\) then \(|[\mathrm{dy} / \mathrm{dx}]|_{(\mathrm{x}=1)}\) \(=\) (a) a cosec a (b) - a cosec a (c) cosec a (d) - cosec a

If \(\mathrm{y}^{2}=\mathrm{p}(\mathrm{x})\) is polynomial function more than 3 degree then \(2(\mathrm{~d} / \mathrm{dx})\left|\mathrm{y}^{3} \mathrm{y}_{2}\right|=\) (a) \(\mathrm{p}^{\prime}(\mathrm{x}) \cdot \mathrm{p}^{\prime \prime}(\mathrm{x})\) (b) \(\mathrm{p}(\mathrm{x}) \cdot \mathrm{p}^{\prime \prime}(\mathrm{x})\) (c) \(\mathrm{p}^{\prime}(\mathrm{x}) \cdot \mathrm{p}^{\prime \prime \prime}(\mathrm{x})\) (d) \(p(x) \cdot p^{\prime \prime \prime}(x)\)

The slope of the tangent at \((2,-1)\) for the curve \(x=t^{2}+3 t-8\) and \(y=2 t^{2}-2 t-5\) is (a) \((6 / 7)\) (b) \(-6\) (c) \((22 / 7)\) (d) \(7 / 6\) )
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