Hello there. To solve this question, we have to remember some properties about polar curves and determining maximum and minimum values.
In this case, we have the function in terms of the angle θ:
[tex]g(\theta)=2\theta-4\sin(\theta)[/tex]
We want to determine its minimum and maximum values on the closed interval:
[tex]\left[0,\,\dfrac{\pi}{2}\right][/tex]
We graph the function as follows:
Notice on the interval, it has a maximum value of 0.
We can determine its minimum value using derivatives, as follows:
[tex]g^{\prime}(\theta)=2-4\cos(\theta)[/tex]
Setting it equal to zero, we obtain
[tex]\begin{gathered} 2-4\cos(\theta)=0 \\ \Rightarrow\cos(\theta)=\dfrac{1}{2} \\ \\ \Rightarrow\theta=\dfrac{\pi}{3} \end{gathered}[/tex]
Taking its second derivative, we obtain
[tex]g^{\prime}^{\prime}(\theta)=4\sin(\theta)[/tex]
And notice that when calculating it on this point, we get
[tex]g^{\prime}^{\prime}\left(\dfrac{\pi}{3}\right)=4\sin\left(\dfrac{\pi}{3}\right)=2\sqrt{3}[/tex]
A positive value, hence it is a minimum point of the function.
Its minimum value is then given by
[tex]g\left(\dfrac{\pi}{3}\right)=2\cdot\dfrac{\pi}{3}-4\sin\left(\dfrac{\pi}{3}\right)=\dfrac{2\pi}{3}-2\sqrt{3}[/tex]
Of course we cannot determine that 0 is a maximum value of this function using derivatives because it is a local maxima on a certain interval, and derivatives can only gives us this value when the slope of the tangent line is equal to zero.
