Question

СРОООЧНО НАДОО найдите точки максимума функции у=-х^2 +25/ х

Answer 1

y'=2x-25/x^2

2x-25/x^2=0

(2x^3-25)/x^2=0

x=(25/2)^(1/3)

 

y''=2+50/x^3

 

y((25/2)^(1/3))=(25/2)^(2/3)+25* (2/25)^(1/3)=(25/2)^(2/3)+1250^(1/3)