Question

How do you find the exact relative maximum and minimum of the polynomial function of `y=2-x^2`?

Answer 1

`y=2-x^2` is a downward opening parabola, with vertex `(0,2)`. Therefore it has a maximum of `2` at `0` and it has no minimum.

Explanation:

If you really want to use calculus, `y'=-2x` so the critical value is `x=0`.

Apply the first derivative test:

`y'` is positive left of `0` and negative right, so `f(0)=2` is a relative maximum.

There are no other critical numbers, so there are no other relative extrema.