At x= 5$\pi$/6 f(x) = 2 sin 3x + 3 cos 3x is ______________ .

Question

KollegeApply · Accepted Answer

none of these \text { Given }: f\left( x \right) = 2 \sin 3x + 3 \cos 3x \Rightarrow f'\left( x \right) = 6 \cos 3x - 9 \sin 3x \text { For a local minima or a local maxima, we must have } f'\left( x \right) = 0 \Rightarrow 6 \cos 3x - 9 \sin 3x = 0 \Rightarrow 6 \cos 3x = 9 \sin 3x \Rightarrow \sin 3x/\cos 3x = $\frac{2}{3}$ \Rightarrow \tan 3x = $\frac{2}{3}$ . . . \left( 1 \right) \text { At x } = 5$\pi$/6: \tan 3x = \tan 5$\pi$/2 \Rightarrow \tan 3x = \tan $\pi$/2 \text { So,} \tan 3x \text { is not defined }. \left[ \tan 3x \neq $\frac{2}{3}$ \text { is not satisfying eq } . \left( 1 \right) \right] \text { Thus, }x = 5$\pi$/6\text { is not a critical point } .

AChoose Your Answer