Expert: Scotto - 4/12/2008

Question
I'm stuck on a calculus problem. I think I'm solving it right but my answer doesn't really seem to make sense.Please could you help me out?

Find the absolute maximum and minimum of the following function on the given interval:
s(t) = ln((t^2) + 1) + 4ln ((t^2)/(t+1)) -4 <= t <= 2

ok so I solved for the derivative and simplified. I arrived at dy/dx =(6t^3 + 8t^2 + 6t + 8)/ (t^4 + t^3 + t^2 + t).

I set this to zero, multiplied the bottom out and then factored getting:
0 = 2[(3t + 4)(t^2 + 1)]

And then t = -4/3 and t = sq. root - 1 which is inadmissable.

I plugged my values (-4/3, -4 and 2) back into the original function s(t) but with the -4/3 being a negative I don't know what to do with it when it comes to ln. For -4 I got 7.44 and 2 I got 0.45, so would these be my max and min?
Also did I solve and factor the derivative correctly or did I go wrong somehow?...
The only thing I can think of where I may not be doing things right is before I took the derivative I split the 4ln part of the equation into 4 ln(t^2) - 4ln(t + 1). Is that right??

Thankyou

Charlotte


Answer
Before taking the derivative, I would simplify the function a little bit.  You have stated how to do this in the bottom of your note by saying that ln(a/b) = ln(a) - ln(b).

The derivative is then 2t/(t^2+1) + 4(2t/t^2) - 4/t.

Multiply the first by t^2/t^2, the second by (t^2+1)/(t^2+1), and the third by t(t^2+1)/(t(t^21)).  Doing this gives
2t^3/*, 4(2t)(t^2+1)/*, and -4t(t^2+1)/*, where * is t^2(t^2+1).

Note that I get a different denominator.

On top, I get 2t^3 + 8t(t^2+1) - (4t^3+4t).  This reduces to
6t^3+4t.  The function is undefined for x<0, so it looks like its increasing and smooth.

Exactly where the first part of your question went wrong, I'm not sure.  I've found it best not to try to look for errors but just take the right answer once you found how to get it.

I won't even try to figure out where this (6t^3 + 8t^2 + 6t + 8)/ (t^4 + t^3 + t^2 + t) cam from.

I hope this has clarified the problem.