find the maximum and minimum values of 2cos2x−cos4x in the range 0<x<π. sketch the curve


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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 bybemath 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 bybemath last updated on 25/Nov/20

Commented byaurpeyz last updated on 25/Nov/20

$t h a n k s$
Commented byaurpeyz 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 bybemath last updated on 25/Nov/20
$let ∅(x)=2cos 2x−cos 4x ∅(x)=2cos 2x−(2cos^2 2x−1) ∅(x)=−2cos^2 2x+2cos 2x+1 ∅(x) = −2(cos^2 2x−cos 2x)+1 ∅(x)=−2[(cos 2x−(1/2))^2 −(1/4)]+1 ∅(x)=−2(cos 2x−(1/2))^2 +(3/2) → { ((max when cos 2x=(1/2))),((min when cos 2x=−1)) :}$
$l e t \emptyset (x) = 2 \cos 2 x - \cos 4 x$ $\emptyset (x) = 2 \cos 2 x - (2 \cos^{2} 2 x - 1)$ $\emptyset (x) = - 2 \cos^{2} 2 x + 2 \cos 2 x + 1$ $\emptyset (x) = - 2 (\cos^{2} 2 x - \cos 2 x) + 1$ $\emptyset (x) = - 2 [{(\cos 2 x - \frac{1}{2})}^{2} - \frac{1}{4}] + 1$ $\emptyset (x) = - 2 {(\cos 2 x - \frac{1}{2})}^{2} + \frac{3}{2}$ $\to {\begin{cases} m a x w h e n \cos 2 x = \frac{1}{2} \\ m i n w h e n \cos 2 x = - 1 \end{cases}$
Commented byMJS_new last updated on 25/Nov/20
$cos 2x =−1+2cos^2 x cos 4x =−1+2cos^2 2x = =−1+2(−1+2cos^2 x)^2 = =1−8cos^2 x +8cos^4 x ⇒2cos 2x −cos 4x = =−3+12cos^2 x −8cos^4 x =f(x) we can always use t=tan (x/2) ⇔ x=2arctan t ⇒ sin x =((2t)/(t^2 +1))∧cos x =−((t^2 −1)/(t^2 +1)) ⇒ −3+12cos^2 x −8cos^4 x = =((t^8 +20t^6 −90t^4 +20t^2 +1)/((t^2 +1)^4 ))=g(t) (dg/dt)=0 −32((t(t^6 −15t^4 +15t^2 −1))/((t^2 +1)^5 ))=0 t(t^6 −15t^4 +15t^2 −1)=0 t_1 =0 t^6 −15t^4 +15t^2 −1=0 trying factors of −1 ⇒ t_(2, 3) =±1 t^4 −14t^2 +1=0 ⇒ t^2 =7±4(√3) ⇒ t=±(√(7±4(√3)))=±(2±(√3)) now we have t∈{−2−(√3), −1, −2+(√3), 0, 2−(√3), 1, 2+(√3)} now t=tan (x/2) ⇒ x=2nπ+2arctan t for 0<x<π we get (x=0∨x=π ⇒ f(x)=1) x=(π/6) ⇒ f(x)=(3/2) x=(π/2) ⇒ f(x)=−3 x=((5π)/6) ⇒ f(x)=(3/2) ⇒ minimum is −3 and maximum is (3/2)$
$\cos 2 x = - 1 + {2 \cos}^{2} x$ $\cos 4 x = - 1 + {2 \cos}^{2} 2 x =$ $= - 1 + 2 {(- 1 + {2 \cos}^{2} x)}^{2} =$ $= 1 - {8 \cos}^{2} x + {8 \cos}^{4} x$ $\Rightarrow 2 \cos 2 x - \cos 4 x =$ $= - 3 + {12 \cos}^{2} x - {8 \cos}^{4} x = f (x)$ $we can always use$ $t = \tan \frac{x}{2} \Leftrightarrow x = 2 \arctan t$ $\Rightarrow \sin x = \frac{2 t}{t^{2} + 1} \land \cos x = - \frac{t^{2} - 1}{t^{2} + 1}$ $\Rightarrow - 3 + {12 \cos}^{2} x - {8 \cos}^{4} x =$ $= \frac{t^{8} + 20 t^{6} - 90 t^{4} + 20 t^{2} + 1}{{(t^{2} + 1)}^{4}} = g (t)$ $\frac{d g}{d t} = 0$ $- 32 \frac{t (t^{6} - 15 t^{4} + 15 t^{2} - 1)}{{(t^{2} + 1)}^{5}} = 0$ $t (t^{6} - 15 t^{4} + 15 t^{2} - 1) = 0$ $t_{1} = 0$ $t^{6} - 15 t^{4} + 15 t^{2} - 1 = 0$ $trying factors of - 1 \Rightarrow t_{2, 3} = \pm 1$ $t^{4} - 14 t^{2} + 1 = 0$ $\Rightarrow t^{2} = 7 \pm 4 \sqrt{3}$ $\Rightarrow t = \pm \sqrt{7 \pm 4 \sqrt{3}} = \pm (2 \pm \sqrt{3})$ $now we have$ $t \in {- 2 - \sqrt{3}, - 1, - 2 + \sqrt{3}, 0, 2 - \sqrt{3}, 1, 2 + \sqrt{3}}$ $now t = \tan \frac{x}{2} \Rightarrow x = 2 n π + 2 \arctan t$ $for 0 < x < π we get$ $(x = 0 \lor x = π \Rightarrow f (x) = 1)$ $x = \frac{π}{6} \Rightarrow f (x) = \frac{3}{2}$ $x = \frac{π}{2} \Rightarrow f (x) = - 3$ $x = \frac{5 π}{6} \Rightarrow f (x) = \frac{3}{2}$ $\Rightarrow minimum is - 3 and maximum is \frac{3}{2}$
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