Thread: Why am I getting stuck?

1. Why am I getting stuck?

This question of my homework was pretty much exactly like the one before it, except with different numbers. Got the previous one, but I'm stuck on this one. Can't get any farther than this:

Find the max and min values of the function s(t) = ln((t2 +1)/(t2-1)) + 6lnt, t is greater than or equal to 1.1 or less than or equal to 10.

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Broke it up using ln rules:

s(t) = ln(t2+1) - ln(t2-1) + 6lnt

s(1.1) = 2.926
s(10) = 13.836

s'(t) = (2t/t2+1) - (2t/t2-1) + 6t

Simplifying gives me this:

s'(t) = (6t4 - 4t2 -6)/(t5-t)

I'm supposed to set it to zero:

0 = 6t4 - 4t2 - 6

But what do I do from there? How do I factor a quartic?

2. Ok

here is what you do if f(t)= $\ln\bigg(\frac{t^2+1}{t^2-1}\bigg)$= $\ln(t^2+1)-\ln(t^2-1)$....you can simplify it further but I see no point...well then f'(t)= $\frac{2t}{t^2+1}-\frac{2t}{t^2-1}$ set that equal to zero and you get t=0...so the other thing you need to check is where f'(t) is undefined...in this case that is at t= $\pm$1...and finally where the original function is undefined on the domain...and that is nowhere...therfore I'd use the critical numbers = $\pm$1 and t=0 to set up the following test intervals (-∞,-1)(-1,0)(0,1)(1,∞) test a value that is an element of each in the first derivative...i will testf'(-2).f'(-.5),f'(.5),f'(2) respectively...we see that there is a sign change at t=-2,t=-.5, and t=.5...there are your max and min points...to find their value plug them back into f(t)

3. What about the "+6lnt" that was part of the question?

You surely must include that, like I did, no?

Have a look at my work... I think it's right but I can go no further.

4. O

haha I am really sorry I completely missed that term...haha if your sure your work is right and you cant use a calculator...well the factoring of that is messy...you could use newtons method...but if all else fails and you dont know high level math(unless you do because then I can tell you an easire way)....use....Intermediate value theorem!....since f'(1) is negative and f'(2) is positive the answer must be between 1 and 2....i then tried 1 and 1.5 it was closer to 0...i tried f'(1.2)=.681...finally i triedf'(1.179)=.003 so it is basically 1.179 I didnt check your work...but if that is the right equation...then thats close...and when I used my calculator's solve function the answer is 1.177

5. 1.177 plugged into the equation gives the value my textbook wants.

Since I never learned any of the things you posted, I don't know how the teacher expects us to solve this.

I'll see him for help on Monday then.

Thanks though!