Question

Найдите точку минимума: y=x^2-18x+40lnx+1

Answer 1

y=x^2-18x+40lnx+1
y'=2x-18+40/x
2x-18+40/x=0
2x^2-18x+40=0
x^2-9x+20=0
x1=5,x2=4
xmin=5

Answer 2

Y ' = 2x - 18 + 40/x

y ' = 0 
2x - 18 + 40/x = 0 
2x^2 - 18x + 40 = 0 /:2
x^2 - 9x + 20 = 0 
D = 81 - 80 = 1
x1 = (9 + 1)/2 = 5
x2 = (9 - 1)/2 = 4

y(4) = 16 - 72 + 55,2 + 1 ≈ 0,2
y(5) = 25 - 90 + 64  + 1 ≈ 0

y min = y (5)