Question

Can you help me find the local extrema and intervals of concavity for g(x)=(x^2+2x-15)/(x^2+2x-3)?

Answer 1

Re-write the derivative please. I forgot ahaha ! :D

Answer 2

g'(x)=24(x+1)/(x+3)^2(x-1)^2 .... do I begin by plugging in all critical points into the original formula to get the local max and min?

Answer 3

For Extrema, only use the critical numbers where derivative is Zero.
that is x = -1

Answer 4

but that's only one number?

Answer 5

I need max AND min

Answer 6

So what, you can't always have both :D
As for y=x^2 You have only one minimum and NO Maximum

Answer 7

Now you have to check where x=-1 is minimum or maximum

Answer 8

and how do I check? like I understand how to get max and min and apparently there is no max in this one... so what i'm asking is how to I check to make sure that i'm right ?

Answer 9

You have to check for second derivative :D
if its positive you have minimum
if its negative you have max

Answer 10

I have to plug -1 into the original equation right? and not into the derivative

Answer 11

What do you mean by take the 2nd derivative>? as in why do I do that

Answer 12

In this case, it is ambiguous that x = -1 is min or max. Right ?

Answer 13

yes

Answer 14

So to remove this ambiguity, second derivative comes to the rescue :D

Answer 15

so I plugged -1 into the original equation of g(x)=x^2+2x-15/x^2+2x-3 and got -4

Answer 16

Yeah the point of extremum is (-1,-4) but you dont know its maxx or min.. Find second derivative to confirm that !

Answer 17

Ok, be right back

Answer 18

g''(x)=-24[3x+6x+7]/(x+3)^3(x-1)^3 is the second derivative

Answer 19

[3x^2+6x+7]

Answer 20

Ok, Plug in -1 :) and tell it's positive or negative

Answer 21

alright, be right back again then haha

Answer 22

undefined?

Answer 23

Hey No !! :/
It is positive.. indicating local minimum and -1

Answer 24

So I may have just calculated it wrong then, eh?

Answer 25

Unfortunately :P Why is it undefined.. :D check again

Answer 26

division by 0

Answer 27

It's -1 not 1 :D There is no Zero in denominator

Answer 28

yeah i've been putting -1 and it keeps coming with the denominator as 0

Answer 29

\[g''(-1) = \frac{-24(3(-1)^2+6(-1)+7)}{(-1+3)^3 (-1-1)^3}\]
Just calculate this now

Answer 30

96?

Answer 31

Whatever be the magnitude.. just check the sign.. Here it is positive right ?

Answer 32

yes

Answer 33

so this would be the max? and the -4 would be the min?

Answer 34

Now how you got 96 ? Why it isnt still undefined ? :D

Answer 35

Since g''(x) is positive.. x=-1 is a MINIMUM for g(x)
Hence there is only one minimum at (-1,-4) and no Maximum for g(x)

Answer 36

alright, makes sense

Answer 37

-4 is the Y-Value of g(x) at x = -1 !! Got it ?
Cheers !

Answer 38

yes, thank you!

Answer 39

i have to go now
Bye See You Later :D
Fan me if you want help from me again ;)

Answer 40

Bye. Thank you again!!!!!