Given the function:
[tex]f(x)=\sqrt{3}\cos x+\sin\text{ x}[/tex]
a)
The first derivative of f(x) is:
[tex]f^{\prime}(x)=-\sqrt{3}\sin\text{ x+}\cos\text{ x}[/tex]
The critical points are found by setting f'(x) = 0:
[tex]-\sqrt{3}\sin\text{ x+}\cos\text{ x=0}[/tex]
Operating:
[tex]\begin{gathered} \cos x=\sqrt{3}\sin\text{ x} \\ \\ \tan x=\frac{1}{\sqrt{3}} \end{gathered}[/tex]
There are two solutions in the interval 0 ≤ x ≤ 2π:
[tex]\begin{gathered} x=\frac{\pi}{6} \\ \\ x=\frac{7\pi}{6} \end{gathered}[/tex]
The second derivative is:
[tex]f^{^^{\prime}^{\prime}}(x)=-\sqrt{3}\cos x\text{-}\sin x[/tex]
Substitute the critical points in the second derivative:
[tex]\begin{gathered} f^{^{\prime\prime}}(\pi/6)=-\sqrt{3}\cos(\pi/6)-\sin(\pi/6) \\ \\ f^{^"}(\pi/6)=-\sqrt{3}\cdot\frac{\sqrt{3}}{2}-\frac{1}{2}=-2 \end{gathered}[/tex]
The second derivative is negative, so this critical point is a maximum.
[tex]\begin{gathered} f^{^{\prime\prime}}(7\pi/6)=-\sqrt{3}\cos(7\pi/6)-\sin(7\pi/6) \\ \\ f^{^"}(7\pi/6)=-\sqrt{3}\cdot\frac{-\sqrt{3}}{2}+\frac{1}{2}=2 \end{gathered}[/tex]
The second derivative is positive, so this critical point is a minimum.
Substituting both points in the original function:
[tex]\begin{gathered} f(\pi/6)=\sqrt{3}\cos(\pi/6)+\sin(\pi/6) \\ \\ f(\pi/6)=3\cdot\frac{\sqrt{3}}{2}+\frac{1}{2}=2 \end{gathered}[/tex]
The maximum is at (π/6, 2).
[tex]\begin{gathered} f(7\pi/6)=\sqrt{3}\cos(7\pi/6)+\sin(7\pi/6) \\ \\ f(7\pi/6)=3\cdot\frac{-\sqrt{3}}{2}-\frac{1}{2}=-2 \end{gathered}[/tex]
The minimum is at (7π/6, -2).
b) The graph is shown below:
Note the function has a maximum when the derivative is 0, then has a minimum when the derivative is 0 again.
The function is decreasing when the derivative is negative and is increasing when the derivative is positive.
The maximum is at (π/6, 2).
The minimum is at (7π/6, -2).
f(π/6) = 2
f(7π/6) = -2
