In the interval [0, 2pi],the slope of the tan | vedclass

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(B) Let $f(x) = e^x \sin x$. The slope of the tangent is given by $f'(x) = e^x \sin x + e^x \cos x = e^x(\sin x + \cos x)$.Let $g(x) = f'(x) = e^x(\si

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$\pi / 2$

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(B) Let $f(x) = e^x \sin x$. The slope of the tangent is given by $f'(x) = e^x \sin x + e^x \cos x = e^x(\sin x + \cos x)$.Let $g(x) = f'(x) = e^x(\sin x + \cos x)$.To find the maximum slope,we find $g'(x) = e^x(\sin x + \cos x) + e^x(\cos x - \sin x) = 2e^x \cos x$.Setting $g'(x) = 0$ for critical points,we get $2e^x \cos x = 0$. Since $e^x 
eq 0$,we have $\cos x = 0$.In the interval $[0, 2\pi]$,$\cos x = 0$ at $x = \pi / 2$ and $x = 3\pi / 2$.Checking the second derivative $g''(x) = 2e^x \cos x - 2e^x \sin x = 2e^x(\cos x - \sin x)$.At $x = \pi / 2$,$g''(\pi / 2) = 2e^{\pi / 2}(0 - 1) = -2e^{\pi / 2} At $x = 3\pi / 2$,$g''(3\pi / 2) = 2e^{3\pi / 2}(0 - (-1)) = 2e^{3\pi / 2} > 0$,which indicates a local minimum.Thus,the slope is maximum at $x = \pi / 2$.

In the interval $[0, 2\pi]$,the slope of the tangent to the function $f(x) = e^x \sin x$ is maximum at $x = \dots$

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