Finding Intervals of Concave Downward with f(x) = 12x^(2/3)

[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)

I can't seem to get the second derivative to work out. Please Help!

 

[dextercioby]

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

If so,then your first derivative is incorrect.

Daniel.

 

[Jameson]

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

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

which simplified gives you

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

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

 

[dextercioby]

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

Would you care to correct your latex graphics in your post...?

Daniel.

 

[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...

 

[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.