Find the critical numbers x1 and x2 of y=(9x^2+3)/4x.
y = (9x^2 + 3)/x
dy/dx = (x(18x) - (9x^2 + 3)/x^2
= (18x^2 - 9x^2 - 3)/x^2
= 0 for critical values
9x^2 = 3
x^2 = 1/3
x = ± 1/√3
dy/dx = (x(18x) - (9x^2 + 3)/x^2
= (18x^2 - 9x^2 - 3)/x^2
= 0 for critical values
9x^2 = 3
x^2 = 1/3
x = ± 1/√3