## soccergal12 3 years ago 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

1. dpaInc

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

2. soccergal12

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

3. dpaInc

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

4. dpaInc

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...

5. dpaInc

or vice versa....

6. soccergal12

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

7. dpaInc

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

8. dpaInc

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

9. dpaInc

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

10. soccergal12

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

11. dpaInc

so what's your critical number(s) ?

12. soccergal12

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

13. dpaInc

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

14. soccergal12

i did that and found the relative max

15. dpaInc

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

16. soccergal12

perfect. thank you :)

17. dpaInc

yw...:)