Posted by **henry** on Friday, November 21, 2008 at 12:38am.

f(x)= x^3/x^2-4

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^2-4)

First find the points where f'(x) = 0

f'(x) = [(x^2-4)(3x^2)-2x^4]/(x^2 -4)^2

(x^2-4)(3x^2)-2x^4 = 0 when f'(x) = 0

3x^4 -12 x^2 -2x^4 = 0

x^2 (x^2-12)

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

## Answer this Question

## Related Questions

- Calculus please help me - f(x) = \frac{ x^3 }{ x^2 - 25 } defined on the ...
- calculus - f(x)= x^3/x^2-16 defined on the interval [-19,16]. Enter points, such...
- Calculus - Answer the following questions for the function f(x)=(x^3-9x^2+27x-27...
- Calculus - f(x)=x*sqrt(x^2+2x+10) + 1*sqrt(x^2+2x+10 defined at the interval [-7...
- math - Answer the following questions for the function f(x)=x(sqrtx2+25) defined...
- Calculus - Suppose that f(x)=ln(3+x^2). Use interval notation to state when f(x...
- Calculus (urgent!!) - Please answer the following questions about the function f...
- Calculus (pleas help!!!) - Please answer the following questions about the ...
- Calculus (pleas help!!!) - Please answer the following questions about the ...
- calculus - Please help me solve. I think increase = concave up and decrease = ...

More Related Questions