Find the intervals of increase and decrease for the following function:

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?

If 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.

Whoops, I combined 2 questions together.

The question is y=x√(4-x).

So 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

To find the intervals of increase and decrease for a function, you need to find the critical points by setting the derivative equal to zero and solving for x.

In this case, the derivative you calculated is:

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

To find the critical points, we set y' equal to zero:

0 = ((4-x)^(-1/2))[((4-x)^-1)-(1/2x)]

Since the factor (4-x)^(-1/2) cannot be zero, we can ignore it and focus on the remaining part of the equation:

((4-x)^-1)-(1/2x) = 0

To simplify it further, we can multiply through by 2x:

2x((4-x)^-1) - 1 = 0

Multiplying through the numerator:

2x - (4-x) = 0

Simplifying:

2x - 4 + x = 0

3x - 4 = 0

Solving for x:

x = 4/3

So the critical point is x = 4/3.

Now, to answer your question about why the zeros of the derivative are not 4 and 0, the mistake lies in your calculation of the derivative. Let's re-calculate the derivative correctly:

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

Notice that, when taking the derivative of the second term, you need to apply the chain rule. The derivative of (1/2x)((4-x)^(-1/2)) is not ((4-x)^(-1/2)), but rather:

- (1/2x^2)((4-x)^(-1/2))

So the corrected derivative is:

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

Now, solving for when y' = 0, you'll find that the critical point is indeed x = 4/3.

To summarize, the corrected critical point for the function is x = 4/3.