Find-the-greatest-and-the-least-values-of-the-function-on-indicates-interval-i-y-sin-x-sin-2x-

Question Number 123396 by bemath last updated on 25/Nov/20

F i n d t h e g r e a t e s t a n d t h e l e a s t v a l u e s

o f t h e f u n c t i o n o n i n d i c a t e s

i n t e r v a l (- \infty, \infty)

(i) y = \sin x \sin 2 x .

Answered by TANMAY PANACEA last updated on 25/Nov/20

y = s i n x \times 2 s i n x c o s x

y = 2 c o s x (1 - {c o s}^{2} x)

y = 2 c - 2 c^{3}

\frac{d y}{d c} = 2 - 6 c^{2}

f o r m a x / m i n \frac{d y}{d c} = 0

6 c^{2} = 2 \to c = \pm \frac{1}{\sqrt{3}}

\frac{d^{2} y}{{d c}^{2}} = - 12 c

{(\frac{d^{2} y}{{d c}^{2}})}_{c = \frac{1}{\sqrt{3}}} = \frac{- 12}{\sqrt{3}} < 0 m a x i m u m

m a x v a l u e y_{m a x} = 2 \times \frac{1}{\sqrt{3}} - 2 \times {(\frac{1}{\sqrt{3}})}^{3}

= \frac{2}{\sqrt{3}} - \frac{2}{3 \sqrt{3}} = \frac{4}{3 \sqrt{3}}

m i n a t c = \frac{- 1}{\sqrt{3}}

y_{m i n} = 2 \times \frac{- 1}{\sqrt{3}} - 2 \times \frac{- 1}{3 \sqrt{3}} = \frac{- 6 + 2}{3 \sqrt{3}} = \frac{- 4}{3 \sqrt{3}}