15
Пошаговое объяснение:
y=7tgx-7x+15
y'=7·(tgx)'-7·x'+15'
y'=7·1/cos²x -7
y'=7·(1/cos²x -1)=7·(1-cos²x)/cos²x=7·sin²x/cos²x=7·tg²x
y'=7·tg²x
7·tg²x=0
tg²x=0
tgx=0
x=π·n, n∈z
Только при n=0, x=0∈[-пи/4);0]
y(-π/4)=7·tg(-π/4)-7·(-π/4)+15=-7+7π/4+15=8+7·π/4
y(0)=7·tg0-7·0+15=-0-0+15=15
Сравним 8+7·π/4
3<π<3,2⇒ 3/4<π/4<3,2/4⇒ 7·3/4<7·π/4<7·3,2/4⇒5,25<7·π/4<5,6⇒
8+5,25<8+7·π/4<8+5,6⇒13,25<8+7·π/4<13,6⇒8+7·π/4<15⇒15- наибольшее значение функции y=7·tgx-7·x+15 на отрезке [-пи/4;0]
ответ:15
1) 9 - 2/3 = 8 1/3
2) 8 1/3 + 6 7/24 = 8 8/24 + 6 7/24 = 14 15/24 = 14 5/8
3) 14 5/8 - 2 1/4 = 14 5/8 - 2 2/8 = 12 3/8
8 13/30 + 13 4/5 - 5 5/6 + 7/10 = 17 1/10
1) 8 13/30 + 13 4/5 = 8 13/30 + 13 24/30 = 21 37/30
2) 21 37/30 - 5 5/6 = 21 37/30 - 5 25/30 = 16 12/30 = 16 4/10
3) 16 4/10 + 7/10 = 16 11/10 = 17 1/10
17 3/4 - 9 1/32 + 4 3/8 - 5 3/16 = 7 29/32
1) 17 3/4 - 9 1/32 = 17 24/32 - 9 1/32 = 8 23/32
2) 8 23/32 + 4 3/8 = 8 23/32 + 4 12/32 = 12 35/32
3) 12 35/32 - 5 3/16 = 12 35/32 - 5 6/32 = 7 29/32
21 4/15 + 1 5/6 - 9 7/30 + 16 7/24 = 30 19/120
1) 21 4/15 + 1 5/6 = 21 8/30 + 1 25/30 = 22 33/30
2) 22 33/30 - 9 7/30 = 13 26/30 = 13 13/15
3) 13 13/15 + 16 7/24 = 13 104/120 + 16 35/120 = 29 139/120 =
= 30 19/120