I have worked out the gradient funtion.
If dy/dx = 8x - 3x^2 , then how you show that dy/dx is postive when
0<x<8/3
To determine if dy/dx is positive when 0 < x < 8/3, we need to analyze the sign of the expression 8x - 3x^2 within this range. Here's how you can show that dy/dx is positive for 0 < x < 8/3:
1. First, find the critical points by setting the expression 8x - 3x^2 equal to zero:
8x - 3x^2 = 0
Factor out an 'x' from the equation:
x(8 - 3x) = 0
Now, solve each factor separately:
x = 0 (from x = 0)
8 - 3x = 0 (from 8 - 3x = 0)
Solving the second factor:
8 - 3x = 0
-3x = -8
x = 8/3
Hence, the critical points are x = 0 and x = 8/3.
2. Now, we need to determine the sign of dy/dx in the range 0 < x < 8/3. To do this, we divide the range into three intervals: x < 0, 0 < x < 8/3, and x > 8/3.
a) For x < 0:
Choose any value within this interval. Let's consider x = -1.
Substitute this value into the expression 8x - 3x^2:
8(-1) - 3(-1)^2 = -8 - 3 = -11
Since the result is negative, dy/dx is negative for x < 0.
b) For 0 < x < 8/3:
Choose any value within this interval. Let's consider x = 1.
Substitute this value into the expression 8x - 3x^2:
8(1) - 3(1)^2 = 8 - 3 = 5
Since the result is positive, dy/dx is positive for 0 < x < 8/3.
c) For x > 8/3:
Choose any value within this interval. Let's consider x = 3.
Substitute this value into the expression 8x - 3x^2:
8(3) - 3(3)^2 = 24 - 27 = -3
Since the result is negative, dy/dx is negative for x > 8/3.
3. Based on the analysis, we can conclude that dy/dx is positive for the range 0 < x < 8/3, which verifies the statement.