Posted by **hawra** on Saturday, March 13, 2010 at 8:35pm.

the function has a real zero in the given interval. approximate this solution correct to two decimal places: f(x)=x^4-x^3-7x^2+5x+10; (2,3)

- pre-calculus -
**Damon**, Saturday, March 13, 2010 at 9:06pm
y =x^4-x^3-7x^2+5x+10

y' = 4x^3 -3x^2 -14x +5

start with x1 = 2.5

calculate y there call it y1

calculate y' there, call it m

next guess at x2 = x1 - y/m

- pre-calculus -
**hawra**, Saturday, March 13, 2010 at 9:19pm
thank you for your time and work. i really appreciate it. god bless you.

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