## BloopityBloop 3 years ago local min/max of f(x)=ln(x^4+8) Did I do this correctly?

1. BloopityBloop

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

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

perfect!

4. asnaseer

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

5. BloopityBloop

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

6. asnaseer

yw :)

7. BloopityBloop

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

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

9. BloopityBloop

ok good. thank you!

10. asnaseer

yw :)