Given y=x²+6 / x, find x if dy/dx = 0
2007-04-16 15:33:56 · 6 answers · asked by th3one101 2 in Science & Mathematics ➔ Mathematics
dy/dx = 2x - 6/x² = 0
2x = 6/x²
x^3 = 3
x = cube root (3).
2007-04-16 15:40:48 · answer #1 · answered by Philo 7 · 0⤊ 0⤋
y = x^2 + 6x^ -1
dy/dx = 2x -6x^ -2
dy/dx = 0 => 2x -6/x^2 =0
=> 2x^3 -6 = 0
=> 2x^3 = 6
=> x^3 = 3
=> x = 3^1/3
2007-04-16 16:02:48 · answer #2 · answered by frank 7 · 0⤊ 0⤋
y=x^2+6/x
therefore,
dy/dx= 2x+D6x^-1
dy/dx=0
2x-6/x^2=0
2x=6/x^2
x=3/x^2
x^3=3
x^3-3=0
by factoring,we can get,
(x-3^1/3)(x^2+x*3^1/3+9^13)
x=3^1/3 or cuberoot of 3
the other factor is disregard because the value of x will be imaginary...
2007-04-16 15:44:31 · answer #3 · answered by king v 1 · 0⤊ 0⤋
dy/dx = 2x-6/x^2=0
2x-6 = 0
x=3
2007-04-16 15:41:02 · answer #4 · answered by ironduke8159 7 · 0⤊ 0⤋
dy/dx = 2x - 6(x^-2) = 0
2x = 6/x^2
x^3 = 3
x = 3^(1/3)
2007-04-16 15:40:55 · answer #5 · answered by metalluka 3 · 0⤊ 0⤋
Give you a hint: let z=y-x => dz/dx=dy/dx-1, z(0)=y(0) Substitute and integrate...
2016-05-17 05:55:07 · answer #6 · answered by kaley 3 · 0⤊ 0⤋