anonymous
  • anonymous
Can you help me find the local extrema and intervals of concavity for g(x)=(x^2+2x-15)/(x^2+2x-3)?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Re-write the derivative please. I forgot ahaha ! :D
anonymous
  • anonymous
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?
anonymous
  • anonymous
For Extrema, only use the critical numbers where derivative is Zero. that is x = -1

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
but that's only one number?
anonymous
  • anonymous
I need max AND min
anonymous
  • anonymous
So what, you can't always have both :D As for y=x^2 You have only one minimum and NO Maximum
anonymous
  • anonymous
Now you have to check where x=-1 is minimum or maximum
anonymous
  • anonymous
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 ?
anonymous
  • anonymous
You have to check for second derivative :D if its positive you have minimum if its negative you have max
anonymous
  • anonymous
I have to plug -1 into the original equation right? and not into the derivative
anonymous
  • anonymous
What do you mean by take the 2nd derivative>? as in why do I do that
anonymous
  • anonymous
In this case, it is ambiguous that x = -1 is min or max. Right ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
So to remove this ambiguity, second derivative comes to the rescue :D
anonymous
  • anonymous
so I plugged -1 into the original equation of g(x)=x^2+2x-15/x^2+2x-3 and got -4
anonymous
  • anonymous
Yeah the point of extremum is (-1,-4) but you dont know its maxx or min.. Find second derivative to confirm that !
anonymous
  • anonymous
Ok, be right back
anonymous
  • anonymous
g''(x)=-24[3x+6x+7]/(x+3)^3(x-1)^3 is the second derivative
anonymous
  • anonymous
[3x^2+6x+7]
anonymous
  • anonymous
Ok, Plug in -1 :) and tell it's positive or negative
anonymous
  • anonymous
alright, be right back again then haha
anonymous
  • anonymous
undefined?
anonymous
  • anonymous
Hey No !! :/ It is positive.. indicating local minimum and -1
anonymous
  • anonymous
So I may have just calculated it wrong then, eh?
anonymous
  • anonymous
Unfortunately :P Why is it undefined.. :D check again
anonymous
  • anonymous
division by 0
anonymous
  • anonymous
It's -1 not 1 :D There is no Zero in denominator
anonymous
  • anonymous
yeah i've been putting -1 and it keeps coming with the denominator as 0
anonymous
  • anonymous
\[g''(-1) = \frac{-24(3(-1)^2+6(-1)+7)}{(-1+3)^3 (-1-1)^3}\] Just calculate this now
anonymous
  • anonymous
96?
anonymous
  • anonymous
Whatever be the magnitude.. just check the sign.. Here it is positive right ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
so this would be the max? and the -4 would be the min?
anonymous
  • anonymous
Now how you got 96 ? Why it isnt still undefined ? :D
anonymous
  • anonymous
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)
anonymous
  • anonymous
alright, makes sense
anonymous
  • anonymous
-4 is the Y-Value of g(x) at x = -1 !! Got it ? Cheers !
anonymous
  • anonymous
yes, thank you!
anonymous
  • anonymous
i have to go now Bye See You Later :D Fan me if you want help from me again ;)
anonymous
  • anonymous
Bye. Thank you again!!!!!

Looking for something else?

Not the answer you are looking for? Search for more explanations.