Question
Limits and derivatives
The point P(16,8) lies on the curve y=√x + 4. If Q is the point (x, √x + 4), find the slope of the secant line PQ for the following value x=16.1
Answers
as x = 16.1 is very close to 16,we can assume slope of secant line PQ is same as that of slope of tangent to the curve at point (16,8)
slope of the tangent is given by derivative
dy/dx = d/dx(√x+4) = 1/(2√x)dy/dx at x=16 is 1/(2*4) = 1/8
Hence required slope = 1/8
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