anonymous
  • anonymous
local min/max of f(x)=ln(x^4+8) Did I do this correctly?
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[f(x)=\ln(x^4+8)\] \[f'(x)=\frac{ 4x^3 }{ x^4+8 }\] Critical point would be 4x^3 = 0 so, x = 0 Next draw number line and check increasing/decreasing intervals: |dw:1352676159858:dw| because \[f'(-1)=\frac{ - }{ + }=negative\] \[f'(1)=\frac{ + }{ + }=positive\]
anonymous
  • anonymous
oh, guess I need to find actual local min and max. So, I use my drawing to say, the local min is 0 and there is no local max for this equation? Is this correct?
asnaseer
  • asnaseer
perfect!

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asnaseer
  • asnaseer
now you just need to find the actual minimum value at x=0
anonymous
  • anonymous
oh, that's what I'm forgetting. Thank you. Gonna figure that out.
asnaseer
  • asnaseer
yw :)
anonymous
  • anonymous
so, plug x=0 into the original function.\[f(0)=\ln(0^4+8)\] \[f(0)=\ln(8)\] and I should've stated the increasing/decreasing intervals. Increasing on (0,infinity). Decreasing on (-infinity,0). How's that?
asnaseer
  • asnaseer
yes - you have it all perfectly correct. :) althought you should specifically mention that the minimum value = f(0) = ln(8)
anonymous
  • anonymous
ok good. thank you!
asnaseer
  • asnaseer
yw :)

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