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

1. BloopityBloop Group Title

$f(x)=\ln(x^4+8)$ $f'(x)=\frac{ 4x^3 }{ x^4+8 }$ 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 Group Title

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 Group Title

perfect!

4. asnaseer Group Title

now you just need to find the actual minimum value at x=0

5. BloopityBloop Group Title

oh, that's what I'm forgetting. Thank you. Gonna figure that out.

6. asnaseer Group Title

yw :)

7. BloopityBloop Group Title

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 Group Title

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

9. BloopityBloop Group Title

ok good. thank you!

10. asnaseer Group Title

yw :)