▶ Refer to Khan academy: Worked example: range of solution curve from slope fieldarrow-up-right
Solve:
Solve: Hint: Try a point or points in each quadrant, like Q1: (2,2), Q2: (-2,2), Q3: (-2,-2), Q4: (2,-2)
Let's only try the point (2, 2), we will get y' = 0, which means the line in Q1 would be horizontal.
(2, 2)
y' = 0
Since 6 is the initial condition, so we make 6 as a boundary.
6
The slope at (0,6) is a negative slope, which seems keeps negative until 4.
(0,6)
4
So the Range would be (4, 6].
(4, 6]
Try out a point, etc, the initial point is (0, 1).
(0, 1)
TAKE DERIVATIVE of each equation, to get the SLOPE of each.
Plug in the x=0 value to get each slope, and eyeball it.
x=0
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