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

Question

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

anonymous · Answer

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

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 · Answer

perfect!

asnaseer · Answer

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

anonymous · Answer

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

asnaseer · Answer

yw :)

anonymous · Answer

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 · Answer

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

anonymous · Answer

ok good.  thank you!

asnaseer · Answer

yw :)