The point in the interval [0, 2pi],where f(x) | vedclass

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(B) To find the point where the slope is maximum,we need to maximize $f'(x)$. Let $g(x) = f'(x) = e^x(\sin x + \cos x)$.For $g(x)$ to have a maximum,w

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$\frac{\pi}{2}$

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(B) To find the point where the slope is maximum,we need to maximize $f'(x)$. Let $g(x) = f'(x) = e^x(\sin x + \cos x)$.For $g(x)$ to have a maximum,we set $g'(x) = f''(x) = 0$.$f'(x) = e^x(\sin x + \cos x)$$f''(x) = e^x(\sin x + \cos x) + e^x(\cos x - \sin x) = 2e^x \cos x$.Setting $f''(x) = 0$ gives $2e^x \cos x = 0$. Since $e^x 
eq 0$,we have $\cos x = 0$.In the interval $[0, 2\pi]$,$\cos x = 0$ at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$.We check the second derivative of $g(x)$,which is $g'(x) = f''(x) = 2e^x \cos x$.$g''(x) = f'''(x) = 2e^x \cos x - 2e^x \sin x = 2e^x(\cos x - \sin x)$.At $x = \frac{\pi}{2}$,$g''(\frac{\pi}{2}) = 2e^{\pi/2}(0 - 1) = -2e^{\pi/2} At $x = \frac{3\pi}{2}$,$g''(\frac{3\pi}{2}) = 2e^{3\pi/2}(0 - (-1)) = 2e^{3\pi/2} > 0$ (Local minimum).Thus,the slope is maximum at $x = \frac{\pi}{2}$.

The point in the interval $[0, 2\pi]$,where $f(x) = e^x \sin x$ has maximum slope,is

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