Originally Posted by

**Linnus** To find the critical points you set the derivative=0 or DNE

I'll do one example and you should know how to do the rest

**Your derivative is -875.01t^2+2250t-1360 (this is the derivative of the velocity right?- since that is what you gave me)**

So set it equal to 0 **and DNE to find the critical points**

-875.01t^2+2250t-1360=0

Use the quadratic equation to find t.

t=.97144573, 1.59999535

-875.01t^2+2250t-1360=DNE does not have a solution which is why I didn't write anything for it

**Critical points of the velocity function exist at t=.97144573, 1.59999535**.

To find the maximum velocity on the interval [1, 1.6) (includes 1 but not 1.6) **This is another way of saying to find the maximum velocity on the interval 1 to 1.6 including 1 but not 1.6**

You would find

V(.97144573)= **plug in .97144573 into the velocity (original function) NOT THE DERIVATIVE **

v(1.59999535)= **plug in 1.59999535**** into the velocity (original function) NOT THE DERIVATIVE **

V(1)= **plug in 1**** into the velocity (original function) NOT THE DERIVATIVE **

And whichever has the largest value out of the 3 is the maximum velocity on that interval.

Is this clear enough?

This is a calculator active problem right? Because using the quadratic equation to find t is a lot of work...