1. ## Max Velocity

Hi guys, I would really appreciate your help with this question.

Determine the max velocity of f(x)=-125(x^2-15x)/x+1

And can this value be found by setting the first derivative of v(t) egual to o(a(t)=o) and solving for t?

I think that the answer for the second question is yes, I just do not know how to solve it.

Here is the first derivative, v(t) -125(x-3)(x+5)/(x+1)^2

And here is the 2nd derivative, a(t) - 4000/(x)^3

ALL THE HELP WOULD BE GREATLY APPRECIATED!!!

2. I think if we eliminated a couple of typing errors and didn't arbitrarily switch between t and x then you would find:

$f(t) =\frac{-125(t^2-15t)}{t+1}$

$v(t)=f'(t) =\frac{-125(t+5)(t-3)}{(t+1)^2}$
$a(t)=v'(t)=\frac{-4000}{(t+1)^3}$

Now the acceleration only approaches 0 as t approaches infinity. As t approaches infinity velocity approaches -125 and this could be a maximum or minimum.

Assuming that we are talking about time and negative time is not allowed then you could have a maximum or minimum at t=0. In fact you do as V(0)=1875

3. Ok,I got all of this. Can the max velocity be 1875 m/s, when max height that I determined is 1125 m? Thats what I though was wrong about my answer