Question

помогите с математикой

Answer 1

y`=28/cos²x -28=0
cos²x=1
(cosx2x+1)/2=1
cos2x+1=2
cos2x=1
2x=0
x=0∈[-π/4;π/4]
y(-π/4)=28*(-1)-28*(-π/4)-7π+7=-21-наим
y(0)=28*0-28*0-7π+7≈-14
y(π/4)=28*1-28*π/4-7π+7=35-14π≈-7-наиб

Answer 2

спасибо большое!:)

Answer 3

y`=28/cos²x -28=0
cos²x=1
(cosx2x+1)/2=1
cos2x+1=2
cos2x=1
2x=0
x=0∈[-π/4;π/4]
y(-π/4)=28*(-1)-28*(-π/4)-7π+7=-21-наим
y(0)=28*0-28*0-7π+7≈-14
y(π/4)=28*1-28*π/4-7π+7=35-14π≈-7-наиб