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December 18, 2014

December 18, 2014

Posted by **Matt** on Saturday, January 29, 2011 at 10:05pm.

- Calc -
**MathMate**, Saturday, January 29, 2011 at 10:16pmIf f(x0)=y0, and f'(x0)=y', then an approximation to f(x0+h)=f(x0)+h*y'

Thus,

x0=125,

h=-1.5

f(x0)=52

y'=4

f(123.5)

=f(125-1.5)

=f(125)+(-1.5)(4)

=52-1.5*4

=46 (approx.)

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