f(x)=\\sin(x)\\cos(x) on (-\\pi,\\pi)\ngive the domain of f using interval notation.\n(-\\pi,\\pi)\nenter…

f(x)=\\sin(x)\\cos(x) on (-\\pi,\\pi)\ngive the domain of f using interval notation.\n(-\\pi,\\pi)\nenter the critical numbers of f, separating with commas if needed. if there are no critical numbers, enter none.\n0,0\nlist the maximal intervals where f is increasing, using interval notation, and separating with commas if needed. if there are no such intervals, enter none.\n\nlist the maximal intervals where f is decreasing, using interval notation, and separating with commas if needed. if there are no such intervals, enter none.\n\nlist the critical points that are a relative maximum. if none are, enter none.\n\nlist the critical points that are a relative minimum. if none are, enter none.

f(x)=\\sin(x)\\cos(x) on (-\\pi,\\pi)\ngive the domain of f using interval notation.\n(-\\pi,\\pi)\nenter the critical numbers of f, separating with commas if needed. if there are no critical numbers, enter none.\n0,0\nlist the maximal intervals where f is increasing, using interval notation, and separating with commas if needed. if there are no such intervals, enter none.\n\nlist the maximal intervals where f is decreasing, using interval notation, and separating with commas if needed. if there are no such intervals, enter none.\n\nlist the critical points that are a relative maximum. if none are, enter none.\n\nlist the critical points that are a relative minimum. if none are, enter none.

Answer

Explanation:

Step1: Simplify the function

Use the double - angle formula ( \sin(2x)=2\sin(x)\cos(x)), so (f(x)=\sin(x)\cos(x)=\frac{1}{2}\sin(2x))

Step2: Find the derivative

Differentiate (y = f(x)=\frac{1}{2}\sin(2x)) using the chain rule. If (u = 2x), then (y=\frac{1}{2}\sin(u)). The derivative (\frac{dy}{du}=\frac{1}{2}\cos(u)) and (\frac{du}{dx}=2). By the chain rule (\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}), so (f^{\prime}(x)=\cos(2x))

Step3: Find critical numbers

Set (f^{\prime}(x) = 0), i.e., (\cos(2x)=0). Then (2x=-\frac{3\pi}{2},-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2}) (since (x\in(-\pi,\pi))). Solving for (x), we get (x =-\frac{3\pi}{4},-\frac{\pi}{4},\frac{\pi}{4},\frac{3\pi}{4})

Step4: Determine intervals of increase and decrease

We use test intervals. Let's consider the intervals ((-\pi,-\frac[Client Connection Error]{3\pi}{4})), ((-\frac{3\pi}{4},-\frac{\pi}{4})), ((-\frac{\pi}{4},\frac{\pi}{4})), ((\frac{\pi}{4},\frac{3\pi}{4})) and ((\frac{3\pi}{4},\pi))

  • For (x\in(-\pi,-\frac{3\pi}{4})), let (x =-\frac{5\pi}{6}), then (f^{\prime}(-\frac{5\pi}{6})=\cos(-\frac{5\pi}{3})=\frac{1}{2}>0)
  • For (x\in(-\frac{3\pi}{4},-\frac{\pi}{4})), let (x =-\frac{\pi}{2}), then (f^{\prime}(-\frac{\pi}{2})=\cos(-\pi)= - 1<0)
  • For (x\in(-\frac{\pi}{4},\frac{\pi}{4})), let (x = 0), then (f^{\prime}(0)=\cos(0)=1>0)
  • For (x\in(\frac{\pi}{4},\frac{3\pi}{4})), let (x=\frac{\pi}{2}), then (f^{\prime}(\frac{\pi}{2})=\cos(\pi)=-1<0)
  • For (x\in(\frac{3\pi}{4},\pi)), let (x=\frac{5\pi}{6}), then (f^{\prime}(\frac{5\pi}{6})=\cos(\frac{5\pi}{3})=\frac{1}{2}>0)

The function (f(x)) is increasing on the intervals ((-\pi,-\frac{3\pi}{4})\cup(-\frac{\pi}{4},\frac{\pi}{4})\cup(\frac{3\pi}{4},\pi)) The function (f(x)) is decreasing on the intervals ((-\frac{3\pi}{4},-\frac{\pi}{4})\cup(\frac{\pi}{4},\frac{3\pi}{4}))

Step5: Find relative maxima and minima

We use the first - derivative test.

  • At (x =-\frac{3\pi}{4}), (f^{\prime}(x)) changes from positive to negative. (f(-\frac{3\pi}{4})=\frac{1}{2}\sin(-\frac{3\pi}{2})=\frac{1}{2})
  • At (x =-\frac{\pi}{4}), (f^{\prime}(x)) changes from negative to positive. (f(-\frac{\pi}{4})=-\frac{1}{2})
  • At (x=\frac{\pi}{4}), (f^{\prime}(x)) changes from positive to negative. (f(\frac{\pi}{4})=\frac{1}{2})
  • At (x=\frac{3\pi}{4}), (f^{\prime}(x)) changes from negative to positive. (f(\frac{3\pi}{4})=-\frac{1}{2})

Answer:

  • Critical numbers: (x =-\frac{3\pi}{4},-\frac{\pi}{4},\frac{\pi}{4},\frac{3\pi}{4})
  • Intervals where (f) is increasing: ((-\pi,-\frac{3\pi}{4})\cup(-\frac{\pi}{4},\frac{\pi}{4})\cup(\frac{3\pi}{4},\pi))
  • Intervals where (f) is decreasing: ((-\frac{3\pi}{4},-\frac{\pi}{4})\cup(\frac{\pi}{4},\frac{3\pi}{4}))
  • Relative maxima: ((-\frac{3\pi}{4},\frac{1}{2}),(\frac{\pi}{4},\frac{1}{2}))
  • Relative minima: ((-\frac{\pi}{4},-\frac{1}{2}),(\frac{3\pi}{4},-\frac{1}{2}))