Homework Help Forum: Calculus

Posted by Z32 on Tuesday, October 27, 2009 at 1:20pm.

F(x) = x^3 / (x-16) is concave up and down on what intervals? I got the derivative to be f' = (x^4 - 48) / (x^2-16)^2 and I found the local min and max but now I need to find where it concaves up and down.

• Calculus - Marth, Tuesday, October 27, 2009 at 2:32pm

  Now you need to take the second derivative. Find where the second derivative is 0 or undefined, and make a sign line around those values. When f''(x)<0, f(x) is concave down; when f''(x)>0, f(x) is concave up.

  
  

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