A few days ago
asymptote, concave regions, inflection point #2?
f(x)= x^3/(x^2-4)
defined on the interval [-15,15]
vertical asymptotes??
concave up on the region ? to ? and ? to ?
inflection point?
Top 1 Answers
A few days ago
Favorite Answer
Vertical asymptotes:
x = -2 and x = 2
To find concavity regions and inflection points, find the second derivative and check its sign:
f’ = (x^4 – 12x^2)/(x^2 – 4)^2
f” = 8x(x^2 + 12)/(x^2 – 4)^3
The curve is concave up <=> f” > 0 or on the intervals (-2, 0) and (2, +infty).
It is concave down on (-infty, -2) and (0, 2).
Iflection points: x=-2, x=0, and x=2.
You can email me if you want more detailed solution.
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