# Calculus:Maximum and Minimum

1. Jan 26, 2005

### thomasrules

I have to find the Max and Minimum in this question.
I'm stuck because I can't find t.

s(t)= 1+2t-8/(t^2+1)

I've got:

s prime(t)= 16t(t^2+1)^-2 +2

Thanks

2. Jan 26, 2005

ok so I assume your equation is $$s(t) = (2t+1) - \frac{8}{t^2+1}$$ Is this correct? Then $$\frac{ds}{dt} = 2 - \frac{16t}{(t^2+1)^2}$$ Then $$t = 0$$

Last edited: Jan 26, 2005
3. Jan 26, 2005

### thomasrules

sorry wrong equation:

(1+2t)-(8/t^2+1)