# Help, Concaved downward?

1. Feb 13, 2005

### ashleyk

Help, Concaved downward??

Let f(x) = 12x^(2/3) Find all the intervals on which f(x) is concaved downward.

I know I have to take the second derivative to find the inflection point to find the interval. I figured the first derivative to be (8-4x^(1/3)/x^(1/3)

2. Feb 13, 2005

### dextercioby

Is this your function:$$f(x)=12x^{\frac{2}{3}}$$ ??

If so,then your first derivative is incorrect.

Daniel.

3. Feb 14, 2005

### Jameson

The first derivative of $$f(x) = 12x^{\frac{2}{3}}$$

is $$(\frac{2}{3})12x^\frac{-1}{3}$$

which simplified gives you

$$8x^\frac{-1}{3}$$

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Sorry for messing up the Latex thing... I'm still new to it

Last edited: Feb 14, 2005
4. Feb 14, 2005

### dextercioby

Daniel.

5. Feb 14, 2005

### ashleyk

sorry the orginial function is wrong...it is actually f(x)= 12x^(2/3)-4x
I still got the derivative of (8-4x^(1/3)/x^(1/3) but I am still having trouble getting through the second derivative...any help would be great...

6. Feb 14, 2005

### dextercioby

Okay,why didn't you leave it in the original handy form...?
$$f'(x)=8x^{-\frac{1}{3}}-4$$

Now differentiate once more...

Daniel.