Question

Find the relative extrema, if any, of the function. Use the Second Derivative Test, if applicable. (If an answer does not exist, enter DNE.)f(x) = 3 sin x + 3 cos x, 0 < x < 2

Answer 1

Given:

Thje critical points of the function are,

`x=(\pi)/(4),(5\pi)/(4)`

Now take the second derivative,

`\begin{gathered} f(x)=3\sin x+3\cos x \\ f^(\prime)(x)=3\cos x-3\sin x \\ f^(\prime\prime)(x)=(d)/(dx)(3\cos x-3\sin x) \\ f^(\prime\prime)(x)=-3\sin x-3\cos x \end{gathered}`

Put the values of the critical points in the second derivatives,

`\begin{gathered} f^(\prime\prime)(x)=-3\sin x-3\cos x \\ f^(\doubleprime)((\pi)/(4))=-3\sin ((\pi)/(4))-3\cos ((\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=}(\pi)/(4) \end{gathered}`

And,

`\begin{gathered} f^(\prime\prime)(x)=-3\sin x-3\cos x \\ f^(\doubleprime)((5\pi)/(4))=-3\sin ((5\pi)/(4))-3\cos ((5\pi)/(4)) \\ =-3(-(√(2))/(2))-3(-(√(2))/(2)) \\ =\frac{6\sqrt[]{2}}{2} \\ =3\sqrt[]{2}>0 \\ \Rightarrow The\text{ function has local minimum at }x=(5\pi)/(4) \end{gathered}`

Answer:

`\begin{gathered} \text{ Maximum (}(\pi)/(4),\: 3√(2)) \\ \text{ Minimum (}(5\pi)/(4),\: -3√(2)) \end{gathered}`