# How do you find the intervals of increasing and decreasing given #y=(2x-8)^(2/3)#?

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To find the intervals of increasing and decreasing for the function ( y = (2x - 8)^{\frac{2}{3}} ), you first need to find its derivative. Then, analyze the sign of the derivative to determine where the function is increasing or decreasing.

Taking the derivative of ( y ) with respect to ( x ), we get: [ \frac{dy}{dx} = \frac{2}{3}(2x - 8)^{-\frac{1}{3}} \times 2 ]

Simplify the derivative: [ \frac{dy}{dx} = \frac{4}{3}(2x - 8)^{-\frac{1}{3}} ]

Now, set the derivative equal to zero to find critical points: [ \frac{4}{3}(2x - 8)^{-\frac{1}{3}} = 0 ]

This implies ( (2x - 8)^{-\frac{1}{3}} = 0 ), but since a real number raised to any power cannot be zero, there are no solutions.

Next, analyze the sign of the derivative in different intervals to determine where the function is increasing or decreasing. Since the derivative is positive when ( (2x - 8)^{-\frac{1}{3}} > 0 ), and negative when ( (2x - 8)^{-\frac{1}{3}} < 0 ), we can conclude:

- ( (2x - 8)^{-\frac{1}{3}} > 0 ) when ( 2x - 8 > 0 ), which gives ( x > 4 ).
- ( (2x - 8)^{-\frac{1}{3}} < 0 ) when ( 2x - 8 < 0 ), which gives ( x < 4 ).

So, the function is increasing for ( x > 4 ) and decreasing for ( x < 4 ). Therefore, the intervals of increasing and decreasing are:

- Increasing: ( x > 4 )
- Decreasing: ( x < 4 )

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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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