can someone please help me with this Calculus question? i would really appreciate it!

Question

for the function f(x)=1/3(x+2)^2(x-3)^2, find the transition points, intervals of increase/decrease,concavity,and asymptotic behavior. 

Answer 1

I would assume the function looks as follows:

 

f(x)=1/(3*(x+2)²(x-3)²). Since we have squares of (x-3) and (x+2), the function is non-negative everywhere.  When x goes to ±∞, function tends to zero, so y=0 is horizontal asymptotic line. Critical points are x=-2 and x=3, those are the points where denominator equals zero. So x=-2 and x=3 are vertical asymptotic lines. 

 

f'(x)=-2/3*(2x-1)/[(x+2)(x-3)]³; f'(x)=0 at x=½. 

f''(x)=10/3*(2x²-2x+3)/[(x+2)(x-3)]⁴;

f''(x) is positive everywhere. The function is concave everywhere.

 

Function increases on the intervals: (-∞, -2); (½, 3)

Function decreases on the intervals: (-2,½); (3,∞)