find the extreme values of the function on the given interval.\n$f(x)=e^{x}\\sin(x)$ on $0,\\pi$\nenter the…

Explanation:

Step1: Find the derivative of (f(x))

Using the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = e^{x}) and (v=\sin(x)). (u^\prime=e^{x}), (v^\prime=\cos(x)) (f^\prime(x)=e^{x}\sin(x)+e^{x}\cos(x)=e^{x}(\sin(x)+\cos(x)))

Step2: Find the critical points

Set (f^\prime(x) = 0), since (e^{x}>0) for all (x), we solve (\sin(x)+\cos(x)=0) (\sin(x)=-\cos(x)), (\tan(x)= - 1) on ([0,\pi]), (x=\frac{3\pi}{4})

Step3: Evaluate the function at critical points and endpoints

• At (x = 0): (f(0)=e^{0}\sin(0)=0)
• At (x=\frac{3\pi}{4}): (f(\frac{3\pi}{4})=e^{\frac{3\pi}{4}}\sin(\frac{3\pi}{4})=\frac{\sqrt{2}}{2}e^{\frac{3\pi}{4}})
• At (x=\pi): (f(\pi)=e^{\pi}\sin(\pi)=0)

Answer:

Maximum value: (\frac{\sqrt{2}}{2}e^{\frac{3\pi}{4}}) Minimum value: (0)