anonymous
  • anonymous
Let g(x)=4x/(x^2 + 1) on the interval [-4,0] . Find the absolute maximum and absolute minimum of g(x) on this interval.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
|dw:1332727333534:dw|
anonymous
  • anonymous
@myininaya
myininaya
  • myininaya
So have you found g' ?

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More answers

anonymous
  • anonymous
@samjordon
myininaya
  • myininaya
Use the quotient rule to find g'
myininaya
  • myininaya
Find critical numbers by setting g'=0 and also finding where g' DNE
anonymous
  • anonymous
g'=0 at 1 and -1
myininaya
  • myininaya
Hint g' exist everywhere since x^2+1 is always greater that 0
anonymous
  • anonymous
at -1 it reaches its absolute minimum
myininaya
  • myininaya
@samjordon please explain how you are getting this. It is against the CoC to just give answers. Thanks.
anonymous
  • anonymous
|dw:1332730291960:dw|
anonymous
  • anonymous
well we have been working thru this already for a while in the chat
myininaya
  • myininaya
\[g'(x)=\frac{(4x)'(x^2+1)-(4x)(x^2+1)'}{(x^2+1)^2}\] I applied the quotient rule: (4x)'=4 (x^2+1)'=2x
anonymous
  • anonymous
so we have reached this point
myininaya
  • myininaya
\[g'(x)=\frac{4(x^2+1)-4x(2x)}{(x^2+1)^2}=\frac{4x^2+4-8x^2}{(x^2+1)^2}=\frac{-4x^2+4}{(x^2+1)^2}\] Ok cool sam :)
myininaya
  • myininaya
Well I will let you guys continue then since I'm not needed
anonymous
  • anonymous
myinaya can u help cuz i a getting confused
anonymous
  • anonymous
this function at -1goes to infinity so can we call that the absolute minimum?
anonymous
  • anonymous
yes?
myininaya
  • myininaya
Ok so like sam said the critical numbers are 1 and -1 good job on that part :) Now we don't care about 1 since 1 is not in the interval [-4,0] All you have to do is compare the values g(0),g(-4),g(-1)
anonymous
  • anonymous
the absolute max is zero right?
anonymous
  • anonymous
i got zero for max
myininaya
  • myininaya
\[g(0)=0; g(-4)=\frac{4(-4)}{(-4)^2+1}=\frac{-16}{17} ; g(-1)=\frac{4(-1)}{(-1)^2+1}=\frac{-4}{2}=-2\] the smallest of these is g(-1) the greatest of these is g(0) => abs min is at x=-1 => abs max is at x=0
anonymous
  • anonymous
ok thank you guys so much i really appreciate your help

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