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BloopityBloop
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local min/max of f(x)=ln(x^4+8)
Did I do this correctly?
 one year ago
 one year ago
BloopityBloop Group Title
local min/max of f(x)=ln(x^4+8) Did I do this correctly?
 one year ago
 one year ago

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BloopityBloop Group TitleBest ResponseYou've already chosen the best response.1
\[f(x)=\ln(x^4+8)\] \[f'(x)=\frac{ 4x^3 }{ x^4+8 }\] Critical point would be 4x^3 = 0 so, x = 0 Next draw number line and check increasing/decreasing intervals: dw:1352676159858:dw because \[f'(1)=\frac{  }{ + }=negative\] \[f'(1)=\frac{ + }{ + }=positive\]
 one year ago

BloopityBloop Group TitleBest ResponseYou've already chosen the best response.1
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?
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
now you just need to find the actual minimum value at x=0
 one year ago

BloopityBloop Group TitleBest ResponseYou've already chosen the best response.1
oh, that's what I'm forgetting. Thank you. Gonna figure that out.
 one year ago

BloopityBloop Group TitleBest ResponseYou've already chosen the best response.1
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?
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
yes  you have it all perfectly correct. :) althought you should specifically mention that the minimum value = f(0) = ln(8)
 one year ago

BloopityBloop Group TitleBest ResponseYou've already chosen the best response.1
ok good. thank you!
 one year ago
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