Kenneth S. answered 11/20/16
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Presumably you know how to interpret the behavior of f' and so the main thing is to differentiate f correctly.
I prefer to use the product rule, differentiating f(x) = -x2(x2-4)-1; let u = -x2 and v = the other factor.
Then u' = -2x and v' = -2x(x2-4)-2 and f' = u'v + uv' = 8x/(x2-4)2.
The critical point is when f' = 0, which occurs if when x=0. BTW, x ≠ ±2 (f & f' undefined)...so there are three critical points.
Here's the relevant numberlline showing CPs and the behaviors of f' and f on the four intervals created by the CPs:
.................(-2)............0...........2....................
f': NEG U NEG 0 POS U POSITIVE ⇐U means undefined
f: decr. decr. incr. incr.
There's a local minimum at x=0, where the function is continuous and switches from decreasing to increasing.
There are vertical asymptotes, x ±2.
The function is decreasing on (-infinity,-2) and (-2,0);
the function is increasing on (0,2) and (2,infinity).
Tom K.
11/20/16