# How do you find the critical points for #y = x/(x-1)^2#?

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To find the critical points for the function ( y = \frac{x}{(x-1)^2} ), we first need to find its derivative. Then, we set the derivative equal to zero and solve for ( x ). Finally, we check these solutions to see if they are valid critical points.

Let's start by finding the derivative of the function ( y ) with respect to ( x ):

[ y = \frac{x}{(x-1)^2} ]

To find the derivative, we can use the quotient rule:

[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} ]

Where ( u = x ) and ( v = (x-1)^2 ). Now, let's find ( u' ) and ( v' ):

[ u' = 1 ] [ v' = 2(x-1) ]

Now, plug these values into the quotient rule formula:

[ y' = \frac{(1)(x-1)^2 - (x)(2(x-1))}{(x-1)^4} ] [ y' = \frac{(x-1)^2 - 2x(x-1)}{(x-1)^4} ] [ y' = \frac{(x^2 - 2x + 1) - (2x^2 - 2x)}{(x-1)^4} ] [ y' = \frac{x^2 - 2x + 1 - 2x^2 + 2x}{(x-1)^4} ] [ y' = \frac{-x^2 + 1}{(x-1)^4} ]

Now, we set the derivative equal to zero and solve for ( x ):

[ \frac{-x^2 + 1}{(x-1)^4} = 0 ]

Since the numerator can only be zero when ( x = \pm 1 ), we only need to check ( x = \pm 1 ) to determine if they are critical points.

Checking ( x = 1 ): [ y' = \frac{-1^2 + 1}{(1-1)^4} = 0 ]

Checking ( x = -1 ): [ y' = \frac{-(-1)^2 + 1}{(-1-1)^4} = \frac{-1 + 1}{(-2)^4} = 0 ]

Therefore, the critical points for the function ( y = \frac{x}{(x-1)^2} ) are ( x = 1 ) and ( x = -1 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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