Question

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

Answer 1

no... nothing wrong there...
what happens when you take the square root of a negative number?

Answer 2

well you can't take a sqaure root of a negative number. so i wouldn't have any inflection points right?

Answer 3

right... that's what that means...

Answer 4

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...

Answer 5

or vice versa....

Answer 6

okay, the question does not ask for concavity, but should i still determine if there is some type of concavity? just to be safe?

Answer 7

sorry... hate when chrome crashes...
no we don't need concavity but the question did ask for points of inflection...

Answer 8

so according to what u got, there are no points of inflection...

Answer 9

now all you need are relative max/mins... do you have your first derivative?

Answer 10

yes , it was (-4x) / (x^2 -4)^2

Answer 11

so what's your critical number(s) ?

Answer 12

0, 2 and -2, i said there was a relatve max at (0, -1/2)

Answer 13

ok.... now do the first derivative test on the intervals:
(-infinity, -2)
(-2, 0)
(0, 2)
(2, infinity)

Answer 14

i did that and found the relative max

Answer 15

oh... sorry... i guess you're done...:)

Answer 16

perfect. thank you :)

Answer 17

yw...:)