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July 31, 2014

July 31, 2014

Posted by **Nevin** on Monday, May 28, 2012 at 9:07pm.

- Calculus -
**Reiny**, Monday, May 28, 2012 at 10:04pmf'(x) = ( a(x^2 - 5x + 4) - (2x-5)(ax+b) )/(x^2 - 5x +4)^2

= 0 for a horizontal line

when x = 2

a(-2) + (2a+b) = 0

-2a + 2a + b = 0

b = 0

also f(2) = -1

2a/(4 - 10 + 4) = -1

2a/-2 = 1

a = 1

a=1 , b=0

so f(x) = x/(x^2 - 5x + 4)

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