Function f(x) = |sin x|,x in left(-frac{pi}{2 | vedclass

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(D) For $x \in \left(-\frac{\pi}{2}, 0\right)$,we have $\sin x < 0$.Therefore,$f(x) = |\sin x| = -\sin x$.To determine the nature of the function,we f

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strictly decreasing function

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(D) For $x \in \left(-\frac{\pi}{2}, 0\right)$,we have $\sin x Therefore,$f(x) = |\sin x| = -\sin x$.To determine the nature of the function,we find its derivative: $f'(x) = \frac{d}{dx}(-\sin x) = -\cos x$.In the interval $\left(-\frac{\pi}{2}, 0\right)$,$\cos x$ is positive (since it is in the fourth quadrant).Thus,$f'(x) = -\cos x Since the derivative is strictly negative,the function $f(x)$ is a strictly decreasing function on the given interval.

Function $f(x) = |\sin x|$,$x \in \left(-\frac{\pi}{2}, 0\right)$ is . . . . . . .

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