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BloopityBloop

  • 3 years ago

local min/max of f(x)=ln(x^4+8) Did I do this correctly?

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  1. BloopityBloop
    • 3 years ago
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    \[f(x)=\ln(x^4+8)\] \[f'(x)=\frac{ 4x^3 }{ x^4+8 }\] Critical point would be 4x^3 = 0 so, x = 0 Next draw number line and check increasing/decreasing intervals: |dw:1352676159858:dw| because \[f'(-1)=\frac{ - }{ + }=negative\] \[f'(1)=\frac{ + }{ + }=positive\]

  2. BloopityBloop
    • 3 years ago
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    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?

  3. asnaseer
    • 3 years ago
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    perfect!

  4. asnaseer
    • 3 years ago
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    now you just need to find the actual minimum value at x=0

  5. BloopityBloop
    • 3 years ago
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    oh, that's what I'm forgetting. Thank you. Gonna figure that out.

  6. asnaseer
    • 3 years ago
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    yw :)

  7. BloopityBloop
    • 3 years ago
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    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?

  8. asnaseer
    • 3 years ago
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    yes - you have it all perfectly correct. :) althought you should specifically mention that the minimum value = f(0) = ln(8)

  9. BloopityBloop
    • 3 years ago
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    ok good. thank you!

  10. asnaseer
    • 3 years ago
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    yw :)

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