Asked by sh
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?
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?
Answers
Answered by
Reiny
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.
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.
Answered by
sh
Whoops, I combined 2 questions together.
The question is y=x√(4-x).
The question is y=x√(4-x).
Answered by
Reiny
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
Mine was -(1/2)(4-x)^(-1/2)(3x-8) = 0
x = 8/3