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January 30, 2015

January 30, 2015

Posted by **kathy** on Wednesday, May 4, 2011 at 7:34pm.

- math -
**Mgraph**, Wednesday, May 4, 2011 at 9:11pmThe equation x^3-4x^2+16=0 has 1 real root

x=-1.68 (approx)

(x^4-4x^3+16x)'=4x^3-12x^2+16=4(x^3-3x^2+4)=4(x+1)(x^2-4x+4)=4(x+1)(x-2)^2

(x^4-4x^3+16x)''=12x^2-24x=12x(x-2)

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