graph the function. then determine for what numbers, if any, the function is discontinuous.\n…

Explanation:

Step1: Recall continuity condition

A function (y = f(x)) is continuous at (x = a) if (\lim_{x\rightarrow a^{-}}f(x)=\lim_{x\rightarrow a^{+}}f(x)=f(a)). We need to check the continuity at (x = 0) and (x=\pi).

Step2: Check continuity at (x = 0)

• Left - hand limit: (\lim_{x\rightarrow0^{-}}f(x)=\lim_{x\rightarrow0^{-}}\cos x=\cos(0) = 1).
• Right - hand limit: (\lim_{x\rightarrow0^{+}}f(x)=\lim_{x\rightarrow0^{+}}\sin x=\sin(0)=0). Since (\lim_{x\rightarrow0^{-}}f(x)\neq\lim_{x\rightarrow0^{+}}f(x)), the function is discontinuous at (x = 0).

Step3: Check continuity at (x=\pi)

• Left - hand limit: (\lim_{x\rightarrow\pi^{-}}f(x)=\lim_{x\rightarrow\pi^{-}}\sin x=\sin(\pi)=0).
• Right - hand limit: (\lim_{x\rightarrow\pi^{+}}f(x)=\lim_{x\rightarrow\pi^{+}}(-\sin x)=-\sin(\pi)=0). And (f(\pi)=-\sin(\pi) = 0). So the function is continuous at (x=\pi).

To graph the function:

• For (y = \cos x) when (-\pi\leq x<0), we know the cosine - function values. (\cos(-\pi)=- 1), (\cos(0) = 1).
• For (y=\sin x) when (0\leq x<\pi), (\sin(0) = 0), (\sin(\frac{\pi}{2})=1), (\sin(\pi)=0).
• For (y =-\sin x) when (\pi\leq x\leq2\pi), (-\sin(\pi)=0), (-\sin(\frac{3\pi}{2}) = 1), (-\sin(2\pi)=0).

Answer:

The function (f(x)) is discontinuous at (x = 0).