Let f(x) = cos left(frac{pi}{x}right), x neq | vedclass

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(A) Given $f(x) = \cos \left(\frac{\pi}{x}\right)$.Taking the derivative with respect to $x$:$f'(x) = -\sin \left(\frac{\pi}{x}\right) \cdot \left(-\f

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$f(x)$ increases in the interval $\left(\frac{1}{2k+1}, \frac{1}{2k}\right)$

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(A) Given $f(x) = \cos \left(\frac{\pi}{x}ight)$.Taking the derivative with respect to $x$:$f'(x) = -\sin \left(\frac{\pi}{x}ight) \cdot \left(-\frac{\pi}{x^2}ight) = \frac{\pi}{x^2} \sin \left(\frac{\pi}{x}ight)$.For $f(x)$ to be increasing, $f'(x) > 0$.Since $\frac{\pi}{x^2} > 0$ for all $x 
eq 0$, we need $\sin \left(\frac{\pi}{x}ight) > 0$.This occurs when $2k\pi Dividing by $\pi$, we get $2k Taking the reciprocal, the inequality reverses: $\frac{1}{2k+1} Thus, $f(x)$ increases in the interval $\left(\frac{1}{2k+1}, \frac{1}{2k}ight)$.For $f(x)$ to be decreasing, $f'(x) This occurs when $(2k+1)\pi Dividing by $\pi$, we get $2k+1 Taking the reciprocal, we get $\frac{1}{2k+2} Thus, $f(x)$ decreases in the interval $\left(\frac{1}{2k+2}, \frac{1}{2k+1}ight)$.

Let $f(x) = \cos \left(\frac{\pi}{x}\right), x \neq 0$. Assuming $k$ is an integer, which of the following is true?

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