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

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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) is positive everywhere. The function is concave everywhere.
Function increases on the intervals: (-∞, -2); (½, 3)
Function decreases on the intervals: (-2,½); (3,∞)