Posted by help on Wednesday, October 27, 2010 at 10:38pm.
Use a linear approximation (or differentials) to estimate the given number
(2.001)^5

calculus  MathMate, Thursday, October 28, 2010 at 10:20am
Let y=x^{5},
y'=5x^{4}.
y(x+h)=y(x)+y'(2)*h (approximately)
Therefore, an approximation to y(2.001) is
y(2.001)
=y(2)+y'(2)*0.001
=2^5 + 5(2)^4*0.001 (approximately)

calculus  sci, Wednesday, December 8, 2010 at 10:13pm
approximate 4 square root of 15.8
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