Find the intervals on which the function is increasing and decreasing.

y= x^(1/4) (x+4)

I took the derivative of the function, it is (1/4)x^(-3/4)

that's all I know.

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- Nov 14th 2008, 06:46 PMSeigihow to find the interval where it's increasing/decreasing?
Find the intervals on which the function is increasing and decreasing.

y= x^(1/4) (x+4)

I took the derivative of the function, it is (1/4)x^(-3/4)

that's all I know. - Nov 14th 2008, 07:14 PMTKHunny
That's not the derivative. What happened to the (x+4)?

Once you find the derivative, where is it positive? That's where the function is increasing. Confusion on this point is not encouraging. It is a fundamental property of the derivative. You would do well to back up and re-read a few chapters. - Nov 14th 2008, 07:32 PMSeigi
i guess i did it wrong, what's the derivative then?

- Nov 14th 2008, 07:36 PMo_O
You tell us:

$\displaystyle \begin{aligned}y & = x^{\frac{1}{4}}(x+4) \\ y & = x^{\frac{5}{4}} + 4x^{\frac{1}{4}} \\ \frac{dy}{dx} & = \frac{d}{dx} \left(x^{\frac{5}{4}} + 4x^{\frac{1}{4}}\right) \end{aligned}$ - Nov 14th 2008, 07:44 PMSeigi
oh, thank you, you set it equal to zero, and find where it is increasing/decreasing right? how do you do that?

- Dec 22nd 2008, 03:10 PMmr fantastic