The number of points in (-infty, infty) for w | vedclass

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(C) Let $f(x) = x^2 - x \sin x - \cos x$.Then,the derivative is $f'(x) = 2x - (\sin x + x \cos x) - (-\sin x) = 2x - x \cos x = x(2 - \cos x)$.Since $

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$2$

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(C) Let $f(x) = x^2 - x \sin x - \cos x$.Then,the derivative is $f'(x) = 2x - (\sin x + x \cos x) - (-\sin x) = 2x - x \cos x = x(2 - \cos x)$.Since $-1 \le \cos x \le 1$,we have $2 - \cos x > 0$ for all real $x$.Thus,$f'(x) = 0$ only at $x = 0$.For $x For $x > 0$,$f'(x) > 0$,so $f(x)$ is strictly increasing.Therefore,$x = 0$ is the point of global minimum.The minimum value is $f(0) = 0^2 - 0 \sin(0) - \cos(0) = -1$.As $x 	o \infty$,$f(x) 	o \infty$,and as $x 	o -\infty$,$f(x) 	o \infty$.Since $f(x)$ is continuous,it crosses the $x$-axis exactly once in $(-\infty, 0)$ and exactly once in $(0, \infty)$.Thus,there are exactly $2$ points where $f(x) = 0$.

The number of points in $(-\infty, \infty)$ for which $x^2 - x \sin x - \cos x = 0$ is

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