Question

what are the critical points of f(x)= (-5x^2)/(2x^2-5)? I already got the f'(x)= (6x^2+10x-15)/ (2x^2-5)^2

Answer 1

okay now to find the [possible critcal poitns we must determine where f' is zero or where it is undefined

Answer 2

what we should do first is try and factor the numertator, did you try that?

Answer 3

yes
i used the quadratic formula and got x=-16+- root (460) / -12

Answer 4

well, then these would be where they could be possible extrema,(although those are really ugly points) what we have to do now is test these points

Answer 5

larange i need your hlp on omy question its hella hard

Answer 6

can you plug these in the calculator and get a numerical value?

Answer 7

yes
i got x= -0.454 and 3.12
approximately

Answer 8

but when i put that into the interval chart
everything is still positive when im trying to find the local max and min

Answer 9

so im thinking im doing something wrong

Answer 10

okay hold on, cause what may be going on is that we are missing an interval or have too many. Where is f undefined and where is f' undefined?

Answer 11

oh VA is +and - root (5/2)

Answer 12

good, those should go into the intervals when testing for exterma

Answer 13

but let me ask you something, did you write the problem correctly, cause the derivative for f you wrote is off

Answer 14

hold on

Answer 15

oh crap -16x^2

Answer 16

-6x^2**

Answer 17

the derivative should be :
\[(50x)/(5-2x^2)^2\]

Answer 18

wait how did you get 50x

Answer 19

okay, so first you factor out the -5

Answer 20

write this down cause it too long to type

Answer 21

1.factor out -5 from orginal function

Answer 22

2. now use the quotient rule on the fraction

Answer 23

ok gonna try this out thxs

Answer 24

and be careful cause its gonna get alittle bit tricky there when using the quotient rule