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soccergal12
find all relative extrema and points of inflection: f(x) = 2 / (x^2 - 4) i got x^2 = -4/3 as an inflection point. but i think i did something wrong
no... nothing wrong there... what happens when you take the square root of a negative number?
well you can't take a sqaure root of a negative number. so i wouldn't have any inflection points right?
right... that's what that means...
but that does not mean the graph is all concave up or all concave down.... there can be places on your graph where in one interval its ccup and in another it's ccdown...
okay, the question does not ask for concavity, but should i still determine if there is some type of concavity? just to be safe?
sorry... hate when chrome crashes... no we don't need concavity but the question did ask for points of inflection...
so according to what u got, there are no points of inflection...
now all you need are relative max/mins... do you have your first derivative?
yes , it was (-4x) / (x^2 -4)^2
so what's your critical number(s) ?
0, 2 and -2, i said there was a relatve max at (0, -1/2)
ok.... now do the first derivative test on the intervals: (-infinity, -2) (-2, 0) (0, 2) (2, infinity)
i did that and found the relative max
oh... sorry... i guess you're done...:)
perfect. thank you :)