Question

Let f(t)=t√(4-t^2) find the abs max and min on interval -1≤t≤2

Answer 1

We are looking for two points, an absolute maximum and minimum. It follows that by being a max and min, the slope at these two points will be zero. Let's take the first derivative of f(x)=t(sqrt(4-t^2) and find where f'(t)=0 If a 0 exists within our interval, it has the potential to be an absolute max or min. I'll let you go from there, but you understand thus far?

Answer 2

I got that part I'm just confused on how to find the max & min