anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
no... nothing wrong there... what happens when you take the square root of a negative number?
anonymous
  • anonymous
well you can't take a sqaure root of a negative number. so i wouldn't have any inflection points right?
anonymous
  • anonymous
right... that's what that means...

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anonymous
  • anonymous
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...
anonymous
  • anonymous
or vice versa....
anonymous
  • anonymous
okay, the question does not ask for concavity, but should i still determine if there is some type of concavity? just to be safe?
anonymous
  • anonymous
sorry... hate when chrome crashes... no we don't need concavity but the question did ask for points of inflection...
anonymous
  • anonymous
so according to what u got, there are no points of inflection...
anonymous
  • anonymous
now all you need are relative max/mins... do you have your first derivative?
anonymous
  • anonymous
yes , it was (-4x) / (x^2 -4)^2
anonymous
  • anonymous
so what's your critical number(s) ?
anonymous
  • anonymous
0, 2 and -2, i said there was a relatve max at (0, -1/2)
anonymous
  • anonymous
ok.... now do the first derivative test on the intervals: (-infinity, -2) (-2, 0) (0, 2) (2, infinity)
anonymous
  • anonymous
i did that and found the relative max
anonymous
  • anonymous
oh... sorry... i guess you're done...:)
anonymous
  • anonymous
perfect. thank you :)
anonymous
  • anonymous
yw...:)

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