find the absolute extrema of the function on the closed interval.\n\n$f(x)=\\sin(x),\\quad\\left\\frac{3\\pi}…

find the absolute extrema of the function on the closed interval.\n\n$f(x)=\\sin(x),\\quad\\left\\frac{3\\pi}{4},\\frac{7\\pi}{4}\\right$\n\nminimum $(x,y)=(\\square)$\n\nmaximum $(x,y)=(\\square)$

find the absolute extrema of the function on the closed interval.\n\n$f(x)=\\sin(x),\\quad\\left\\frac{3\\pi}{4},\\frac{7\\pi}{4}\\right$\n\nminimum $(x,y)=(\\square)$\n\nmaximum $(x,y)=(\\square)$

Answer

Explanation:

Step1: Find the derivative of the function

The derivative of (y = f(x)=\sin(x)) is (f^{\prime}(x)=\cos(x)). Set (f^{\prime}(x) = 0), so (\cos(x)=0). In the interval (\left[\frac{3\pi}{4},\frac{7\pi}{4}\right]), (x=\frac{3\pi}{2}) (since (\cos(x) = 0) when (x=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}) and (\frac{3\pi}{2}\in\left[\frac{3\pi}{4},\frac{7\pi}{4}\right])).

Step2: Evaluate the function at critical points and endpoints

  • Evaluate (f(x)) at (x = \frac{3\pi}{4}): (f\left(\frac{3\pi}{4}\right)=\sin\left(\frac{3\pi}{4}\right)=\frac{\sqrt{2}}{2})
  • Evaluate (f(x)) at (x=\frac{3\pi}{2}): (f\left(\frac{3\pi}{2}\right)=\sin\left(\frac{3\pi}{2}\right)=- 1)
  • Evaluate (f(x)) at (x=\frac{7\pi}{4}): (f\left(\frac{7\pi}{4}\right)=\sin\left(\frac{7\pi}{4}\right)=-\frac{\sqrt{2}}{2})

Answer:

minimum ((x,y)=\left(\frac{3\pi}{2},-1\right)) maximum ((x,y)=\left(\frac{3\pi}{4},\frac{\sqrt{2}}{2}\right))