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

  • Thread starter Thread starter ashleyk
  • Start date Start date
ashleyk
Messages
22
Reaction score
0
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!
 
Physics news on Phys.org
Is this your function:f(x)=12x^{\frac{2}{3}} ??

If so,then your first derivative is incorrect.

Daniel.
 
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
 
Last edited by a moderator:
Jameson said:
The first derivative of f(x) = 12x^{\frac{2}{3}}

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

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

Now differentiate once more...

Daniel.
 
Back
Top