If dy/dx = 8x - 3x^2 , then how you show that dy/dx is postive when
0<x<8/3
To show that dy/dx is positive when 0 < x < 8/3, we can use the given equation dy/dx = 8x - 3x^2 and analyze its behavior within the given interval.
1. Identify the critical points:
Critical points are the values of x where the derivative could change sign. In this case, we focus on 0 < x < 8/3. To find the critical points, we set dy/dx = 0 and solve for x:
8x - 3x^2 = 0
Factor out x:
x(8 - 3x) = 0
This equation is satisfied when x = 0 or 8 - 3x = 0. From the second equation, we get 8 = 3x, so x = 8/3. Thus, the critical points are x = 0 and x = 8/3.
2. Examine the signs of dy/dx for each interval:
Now, we divide the interval 0 < x < 8/3 into three segments: (0, 8/3/2), (8/3/2, 8/3), and (8/3, ∞). We test a value within each segment to determine the sign of dy/dx. Let's choose the midpoint of each interval and evaluate dy/dx at these points.
For the interval (0, 8/3/2) with x = 1, we substitute x = 1 into the given equation:
dy/dx = 8(1) - 3(1^2) = 8 - 3 = 5
Since dy/dx > 0 at this point, the derivative is positive in this interval.
For the interval (8/3/2, 8/3) with x = 5/3, we substitute x = 5/3:
dy/dx = 8(5/3) - 3(5/3)^2 = 40/3 - 25/3 = 15/3 = 5
Again, dy/dx > 0 at this point, indicating a positive derivative in this interval as well.
Lastly, for the interval (8/3, ∞), we choose x = 3. The substitution gives us:
dy/dx = 8(3) - 3(3^2) = 24 - 27 = -3
In this interval, dy/dx < 0, indicating a negative derivative.
Since dy/dx is positive in the intervals (0, 8/3/2) and (8/3/2, 8/3), we have shown that dy/dx is positive when 0 < x < 8/3.