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December 17, 2014

December 17, 2014

Posted by **sh** on Sunday, December 13, 2009 at 9:56pm.

h(x)=(x^3)(x-1)^4

h'(x)=[(3x^2)(x-1)^4]+(4x^3)(x-1)^3

How do I find the zeros?

Thanks.

- Calculus -
**Reiny**, Sunday, December 13, 2009 at 10:07pmThe zeros of the derivative or the zeros of the function ?

I will assume you want the zeros of the derivative.

factor it first

= x^2(x-1)^3[4x + 3(x-1)]

= x^2(x-1)^3(7x-3)

we set this equal to zero to get

x = 0, x = 1 and x = 3/7

remember that the function increases when the first derivative is positive, and decreases when that derivative is negative.

Can you take it from here ?

- Calculus -
**sh**, Sunday, December 13, 2009 at 10:09pmDo you factor it by taking the lowest ^ ? I can continue from there, thanks! :)

- Calculus -
**sh**, Sunday, December 13, 2009 at 10:21pmThe answer key indicated that 0 is not a zero.

- Calculus -
**Reiny**, Sunday, December 13, 2009 at 10:29pmYour question was,

"how do I find the zeros"

x = 0 is a solution to h(x) = 0 and h'(x) = 0

- Calculus -
**sh**, Sunday, December 13, 2009 at 10:35pmIf x=0, shouldn't it be used to determine the increase and decrease?

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