f(x) = 3·ln(2x) - x/3 , x > 0.
Here find the derivetive's 0 point
2007-12-13 19:29:06 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
y = 3 ln (2x) - x/3
dy/dx = f `(x) = (1/x)(3) - 1/3
dy/dx = f `(x) = 3/x - 1/3 = 0
3/x = 1/3
x = 9
2007-12-14 00:21:09 · answer #1 · answered by Como 7 · 3⤊ 0⤋
f(x) = 3·ln(2x) - x/3 , x > 0
=> f'(x) = 3*2*(1/2x) - 1/3
f'(x) = 0
=> 3/x - 1/3 = 0
=> 3/x = 1/3
=> x = 9.
2007-12-14 03:41:27 · answer #2 · answered by Madhukar 7 · 0⤊ 0⤋
Why don't you try calculating the derivative first?
f '(x) = 3/x - 1/3 .
Then, where is this zero?
You can then solve
f '(x) = 3/x - 1/3 = 0.
2007-12-14 03:32:58 · answer #3 · answered by nicholasm40 3 · 0⤊ 1⤋
f(x) = 3 - ln(2x) - x/3
dx/dy = 0 - (1/2x)(2) - 1/3
dx/dy = - 1/x - 1/3
For the zero value, let the derivate be equal to zero.
dx/dy = - 1/x - 1/3
0 = - 1/x - 1/3
1/3 = - 1/x
3 = - x
-3 = x
If you mean:
f(x) = 3ln(2x) - x/3
dy/dx = 3(1/2x)(2) - 1/3
2007-12-14 03:53:23 · answer #4 · answered by Sparks 6 · 0⤊ 0⤋