# how to find the interval where it's increasing/decreasing?

• Nov 14th 2008, 06:46 PM
Seigi
how 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 PM
TKHunny
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 PM
Seigi
i guess i did it wrong, what's the derivative then?
• Nov 14th 2008, 07:36 PM
o_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 PM
Seigi
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 PM
mr fantastic
Quote:

Originally Posted by Seigi
oh, thank you, you set it equal to zero, and find where it is increasing/decreasing right? how do you do that?

Use the solutions to dy/dx = 0 to solve dy/dx > 0 and dy/dx < 0.

It would help if you showed some working. As it stands, you haven't even shown the equations and inequations you're trying to solve, let alone where you get stuck in trying to solve them.