Posted by henry on Friday, November 21, 2008 at 12:38am.
f(x)= x^3/x^24
defined on the interval [18,20]
f(x) is concave up on the region ? to ? and ? to ?
the inflection points are ?, ? and ?

calculus!!  drwls, Friday, November 21, 2008 at 6:49am
I assume you mean f(x) = x^3/(x^24)
First find the points where f'(x) = 0
f'(x) = [(x^24)(3x^2)2x^4]/(x^2 4)^2
(x^24)(3x^2)2x^4 = 0 when f'(x) = 0
3x^4 12 x^2 2x^4 = 0
x^2 (x^212)
x = 0, +2sqrt3 and 2sqrt3
Now evaluate the second derivative at those three critical points. The function is concave up if f''(x) > 0 and is an inflection point if f''(x) = 0
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