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January 30, 2015

January 30, 2015

Posted by **sh** on Sunday, December 13, 2009 at 10:33pm.

y=x√((x-1)^4)

y'=[(4-x)^(1/2)]+(1/2)((4-x)^(-1/2))(-1)(x)

y'=[(4-x)^(1/2)]-(1/2x)((4-x)^(-1/2))

y'=((4-x)^(-1/2))[((4-x)^-1)-1/2x]

I've already factored it, how come the zeros of the derivative aren't 4 and 0?

- Calculus -
**Reiny**, Sunday, December 13, 2009 at 10:48pmIf I interpret your first line correctly it is

y = x√((x-1)^4)

Isn't that

y = x(x-1)^2 ??

then

y' x(2)(x-1) + (x-1)^2

= (x-1)(2x - x + 1)

= (x-1)(x+1)

= 0

so x = 1 or x = -1 are the zeros of the derivative.

I don't see where the (4-x)'s in your second line come from, there wasn't even a 4-x in the equation.

- Calculus -
**sh**, Sunday, December 13, 2009 at 10:52pmWhoops, I combined 2 questions together.

The question is y=x√(4-x).

- Calculus -
**Reiny**, Sunday, December 13, 2009 at 11:14pmSo what did you get for you factored derivative set equal to zero ?

Mine was -(1/2)(4-x)^(-1/2)(3x-8) = 0

x = 8/3

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