Найдите наибольшее значение функции f(x)=3(2x-4)^4-(2x-4)^5. при |х-2|< =1. : -)
173
321
Ответы на вопрос:
/x-2/≤1⇒-1≤x-2≤1⇒1≤x≤3⇒x∈[1; 3] f`(x)=24(2x-4)³-10(2x-4)^4=2(2x-4)³(12-10x+20)=2(2x-4)³(22-10x)=0 2x-4=0⇒x=2∈[1; 3] 22-10x=0⇒x=2,2∈[1; 3] y(1)=3*16+32=48+32=80-наиб y(2)=3*0-0=0-наим y(2,2)=3*0,0256-0,01024=0.0768-0,01024=0,06656 y(3)=3*16-32=48-32=16
1) 8(1-sin²x) + 6sinx = 3 8 - 8sin²x + 6sinx -3 = 0 8sin²x -6sinx -5 = 0 решаем как квадратное d = 36 -4*8*(-5) = 196 sinx = (6+14)/16 = 20/16 ( нет решений) sinx =(6 -14)/16 = -1/2 sinx = -1/2 x = (-1)^(n+1)π/6 + nπ, n ∈z 2)cos²2x + cos6x -sin²2x = 0 cos4x + cos6x = 0 ( формула суммы косинусов) 2сos5xcosx = 0 cos5x = 0 или cosx = 0 5x = π/2 + πk , k ∈z x = π/2 + πn , n ∈z x = π/10 + πk/5, k ∈z 3) (cos²2x - sin²2x)(cos²2x+sin²2x) = √3/2 cos²2x -sin²2x = √3/2 cos4x = √3/2 4x = +-arccos(√3/2) + 2πk , k ∈z 4x = +-π/6 +2πk , k ∈z x = +-π/24 + πk/2 , k ∈z 4) 4sin²x -8sinxcosx +10cos²x = 3*1 4sin²x -8sinxcosx +10cos²x = 3(sin²x + cos²x) 4sin²x -8sinxcosx +10cos²x -3sin²x - 3cos²x = 0 sin²x -8sinxcosx +7cos²x = 0 | : cos²x tg²x - 8tgx +7 = 0 по т. виета tgx = 1 или tgx = 7 x = π/4 + πk , k ∈z x = arctg7 + πn , n ∈z 5) 1 + cosx + cos2x = 0 1 + cosx + 2cos²x - 1 = 0 cosx + 2cos²x = 0 cosx(1 +2cosx) = 0 cosx = 0 или 1 + 2cosx = 0 x = π/2 + πk , k ∈z cosx = -1/2 х = +-arccos(-1/2) +2πn , n ∈z x = +-2π/3 + 2πn , n ∈z 6) -cosx > -0,5 cosx < 0,5 -π/3 + 2πk < x < π/3 + 2πk , k ∈z
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