Respuesta :
Given:
[tex]f(x)=3\sin x+3\cos x,0To find the relative maxima of the given function use second derivative test,[tex]\begin{gathered} \text{Differentiate f(x) with respect to x,} \\ f^{\prime}(x)=\frac{d}{dx}(3\sin x+3\cos x) \\ f^{\prime}(x)=3\cos x-3\sin x \\ \text{Set }f^{\prime}(x)=0 \\ 3\cos x-3\sin x=0 \\ \cos x=\sin x \\ \frac{\sin x}{\cos x}=1 \\ \tan x=1 \\ x=\tan ^{-1}(1) \\ x=\frac{\pi}{4}+\pi n \\ \text{But for 0}<\text{x<2}\pi \\ \text{x will take the values}, \\ x=\frac{\pi}{4},\frac{5\pi}{4}\in(0,2\pi) \end{gathered}[/tex]Thje critical points of the function are,
[tex]x=\frac{\pi}{4},\frac{5\pi}{4}[/tex]Now take the second derivative,
[tex]\begin{gathered} f(x)=3\sin x+3\cos x \\ f^{\prime}(x)=3\cos x-3\sin x \\ f^{\prime\prime}(x)=\frac{d}{dx}(3\cos x-3\sin x) \\ f^{\prime\prime}(x)=-3\sin x-3\cos x \end{gathered}[/tex]Put the values of the critical points in the second derivatives,
[tex]\begin{gathered} f^{\prime\prime}(x)=-3\sin x-3\cos x \\ f^{\doubleprime}(\frac{\pi}{4})=-3\sin (\frac{\pi}{4})-3\cos (\frac{\pi}{4}) \\ =-\frac{3}{\sqrt[]{2}}-\frac{3}{\sqrt[]{2}} \\ =\frac{-6}{\sqrt[]{2}} \\ =-3\sqrt[]{2}<0 \\ \Rightarrow Function\text{ has local maximum at x=}\frac{\pi}{4} \end{gathered}[/tex]And,
[tex]\begin{gathered} f^{\prime\prime}(x)=-3\sin x-3\cos x \\ f^{\doubleprime}(\frac{5\pi}{4})=-3\sin (\frac{5\pi}{4})-3\cos (\frac{5\pi}{4}) \\ =-3(-\frac{\sqrt{2}}{2})-3(-\frac{\sqrt{2}}{2}) \\ =\frac{6\sqrt[]{2}}{2} \\ =3\sqrt[]{2}>0 \\ \Rightarrow The\text{ function has local minimum at }x=\frac{5\pi}{4} \end{gathered}[/tex]Answer:
[tex]\begin{gathered} \text{ Maximum (}\frac{\pi}{4},\: 3\sqrt{2}) \\ \text{ Minimum (}\frac{5\pi}{4},\: -3\sqrt{2}) \end{gathered}[/tex]