Given-f-x-2-sin-2-x-sin-x-1-0-x-2pi-Find-maximum-and-minumum-value-of-f-x-without-differential-

Question Number 87171 by naka3546 last updated on 03/Apr/20

G i v e n f (x) = 2 \sin^{2} x - \sin x + 1, 0 ⩽ x ⩽ 2 π

F i n d m a x i m u m a n d m i n u m u m v a l u e

o f f (x) w i t h o u t d i f f e r e n t i a l .

Commented by john santu last updated on 03/Apr/20

for \sin x = 1 \Rightarrow y_{1} = 2 - 1 + 1 = 2

for \sin x = - 1 \Rightarrow y_{2} = 2 + 1 + 1 = 4

for \sin x = \frac{1}{4} \Rightarrow y_{3} = \frac{1}{8} - \frac{1}{4} + 1

y_{3} = \frac{1}{8} + \frac{3}{4} = \frac{7}{8}

\max = 4 & \min = \frac{7}{8}

Commented by mr W last updated on 03/Apr/20

f (x) = 2 (\sin^{2} x - 2 \times \frac{1}{4} \sin x + \frac{1}{4^{2}}) + 1 - 2 \times \frac{1}{4^{2}}

= 2 {(\sin x - \frac{1}{4})}^{2} + \frac{7}{8}

m i n . = \frac{7}{8}

m a x . = 2 {(- 1 - \frac{1}{4})}^{2} + \frac{7}{8} = 4

Facebook Tweet Pin