find-the-maximum-and-minimum-values-of-2cos2x-cos4x-in-the-range-0-lt-x-lt-pi-sketch-the-curve-

Question Number 123337 by aurpeyz last updated on 25/Nov/20

f i n d t h e m a x i m u m a n d m i n i m u m

v a l u e s o f 2 c o s 2 x - c o s 4 x i n t h e r a n g e

0 < x < π . s k e t c h t h e c u r v e

Commented by bemath last updated on 25/Nov/20

m a x = 1.5; m i n = - 3

r a n g e \Rightarrow - 3 ⩽ y ⩽ 1.5

Commented by bemath last updated on 25/Nov/20

Commented by aurpeyz last updated on 25/Nov/20

t h a n k s

Commented by aurpeyz last updated on 25/Nov/20

p l s e x p l a i n h o w y o u g o t m i n = - 3 .

i t r i e d s o h a r d b u t c o u l d n t

Commented by bemath last updated on 25/Nov/20

l e t \emptyset (x) = 2 \cos 2 x - \cos 4 x

\emptyset (x) = 2 \cos 2 x - (2 \cos^{2} 2 x - 1)

\emptyset (x) = - 2 \cos^{2} 2 x + 2 \cos 2 x + 1

\emptyset (x) = - 2 (\cos^{2} 2 x - \cos 2 x) + 1

\emptyset (x) = - 2 [{(\cos 2 x - \frac{1}{2})}^{2} - \frac{1}{4}] + 1

\emptyset (x) = - 2 {(\cos 2 x - \frac{1}{2})}^{2} + \frac{3}{2}

\to {\begin{cases} m a x w h e n \cos 2 x = \frac{1}{2} \\ m i n w h e n \cos 2 x = - 1 \end{cases}

Commented by MJS_new last updated on 25/Nov/20

\cos 2 x = - 1 + {2 \cos}^{2} x

\cos 4 x = - 1 + {2 \cos}^{2} 2 x =

= - 1 + 2 {(- 1 + {2 \cos}^{2} x)}^{2} =

= 1 - {8 \cos}^{2} x + {8 \cos}^{4} x

\Rightarrow 2 \cos 2 x - \cos 4 x =

= - 3 + {12 \cos}^{2} x - {8 \cos}^{4} x = f (x)

we can always use

t = \tan \frac{x}{2} \Leftrightarrow x = 2 \arctan t

\Rightarrow \sin x = \frac{2 t}{t^{2} + 1} \land \cos x = - \frac{t^{2} - 1}{t^{2} + 1}

\Rightarrow - 3 + {12 \cos}^{2} x - {8 \cos}^{4} x =

= \frac{t^{8} + 20 t^{6} - 90 t^{4} + 20 t^{2} + 1}{{(t^{2} + 1)}^{4}} = g (t)

\frac{d g}{d t} = 0

- 32 \frac{t (t^{6} - 15 t^{4} + 15 t^{2} - 1)}{{(t^{2} + 1)}^{5}} = 0

t (t^{6} - 15 t^{4} + 15 t^{2} - 1) = 0

t_{1} = 0

t^{6} - 15 t^{4} + 15 t^{2} - 1 = 0

trying factors of - 1 \Rightarrow t_{2, 3} = \pm 1

t^{4} - 14 t^{2} + 1 = 0

\Rightarrow t^{2} = 7 \pm 4 \sqrt{3}

\Rightarrow t = \pm \sqrt{7 \pm 4 \sqrt{3}} = \pm (2 \pm \sqrt{3})

now we have

t \in {- 2 - \sqrt{3}, - 1, - 2 + \sqrt{3}, 0, 2 - \sqrt{3}, 1, 2 + \sqrt{3}}

now t = \tan \frac{x}{2} \Rightarrow x = 2 n π + 2 \arctan t

for 0 < x < π we get

(x = 0 \lor x = π \Rightarrow f (x) = 1)

x = \frac{π}{6} \Rightarrow f (x) = \frac{3}{2}

x = \frac{π}{2} \Rightarrow f (x) = - 3

x = \frac{5 π}{6} \Rightarrow f (x) = \frac{3}{2}

\Rightarrow minimum is - 3 and maximum is \frac{3}{2}

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