Question

Найдите наибольшее значение функции y = 3 корень 2 cos x + 3x - 3п/4 +7 на отрезке [0;п/2]

Answer 1

Y`=-3√2sinx+3=0
sinx=1/√2
x=π/4∈[0;π/2]
x=3π/4∉[0;π/2]
y(0)=3√2*1-3*0-3π/4+7≈4,2-2,4+7≈8,8
y(π/4)=3√2*1/√2+3π/4-3π/4+7=10 наиб
y(π/2)=3√3*0+3π/2-3π/4+7=3π/4+7≈2,4+7=9,4