Question:

Step 1: Simplify the given function using the double angle formula. The identity $2\sin\theta\cos\theta = \sin 2\theta$ is applied with $\theta = \frac{\pi}{4} - \frac{x}{2}$. $$y = 2e^x \sin\left(\frac{\pi}{4} - \frac{x}{2}\right)\cos\left(\frac{\pi}{4} - \frac{x}{2}\right)$$ $$y = e^x \sin\left(2\left(\frac{\pi}{4} - \frac{x}{2}\right)\right)$$ $$y = e^x \sin\left(\frac{\pi}{2} - x\right)$$ $$y = e^x \cos x$$ Step 2: Determine the slope by computing the derivative. The slope of the tangent is given by $\frac{dy}{dx}$. Using the product rule, $\frac{d}{dx}(uv) = u'v + uv'$: $$\frac{dy}{dx} = \frac{d}{dx}(e^x \cos x) = (e^x)(\cos x) + (e^x)(-\sin x)$$ $$\frac{dy}{dx} = e^x(\cos x - \sin x)$$ Step 3: Find the critical points of the slope function. Let $g(x) = e^x(\cos x - \sin x)$ represent the slope. To find where the slope is minimum, we compute the derivative of $g(x)$, denoted as $g'(x)$, using the product rule: $$g'(x) = \frac{d}{dx}(e^x(\cos x - \sin x))$$ $$g'(x) = (e^x)(\cos x - \sin x) + (e^x)(-\sin x - \cos x)$$ $$g'(x) = e^x(\cos x - \sin x - \sin x - \cos x)$$ $$g'(x) = e^x(-2\sin x)$$ $$g'(x) = -2e^x \sin x$$ Step 4: Analyze the critical points to determine the minimum slope. Set $g'(x) = 0$ to find critical points: $$-2e^x \sin x = 0$$ Since $e^x > 0$ for all $x$, we must have $\sin x = 0$. In the interval $0 \leq x \leq 2\pi$, the solutions are $x = 0, \pi, 2\pi$. Now, analyze the sign of $g'(x) = -2e^x \sin x$: * For $x \in (0, \pi)$: $\sin x > 0$, so $g'(x) = -2e^x(\text{positive value}) < 0$. This indicates that $g(x)$ is decreasing. * For $x \in (\pi, 2\pi)$: $\sin x < 0$, so $g'(x) = -2e^x(\text{negative value}) > 0$. This indicates that $g(x)$ is increasing. The change in sign of $g'(x)$ from negative to positive at $x = \pi$ confirms that $g(x)$ has a local minimum at $x = \pi$. To confirm the global minimum within the interval $[0, 2\pi]$, evaluate $g(x)$ at the critical points and endpoints: * At $x = 0$: $g(0) = e^0(\cos 0 - \sin 0) = 1(1 - 0) = 1$. * At $x = \pi$: $g(\pi) = e^\pi(\cos \pi - \sin \pi) = e^\pi(-1 - 0) = -e^\pi$. * At $x = 2\pi$: $g(2\pi) = e^{2\pi}(\cos(2\pi) - \sin(2\pi)) = e^{2\pi}(1 - 0) = e^{2\pi}$. Comparing the values $1$, $-e^\pi$, and $e^{2\pi}$, the minimum value is $-e^\pi$. The slope of the tangent to the curve is minimum at $x = \pi$.