Question

Over the x-value interval `[−10,10]`, what are the absolute extrema of `f(x)=x^2`?

Answer 1

The absolute maximum value is 100 at `x=\pm 10` and the absolute minimum value is 0 at `x=0`.

Explanation:

This is clear just by knowing the graph of `y=f(x)=x^2` as an upward-opening parabola with a vertex at the point `(0,0)`.

You can also use calculus (just a little preview if you are in precalculus). Since the derivative `f'(x)=2x`, the only critical point of the function is `x=0`. Now you compare the value of the continuous function `f` at the critical point and the endpoints `x=\pm 10` of the closed and bounded interval `[-10,10]`:

`f(0)=0^2=0`, `f(\pm 10)=10^2=100`.

This guarantees that the absolute maximum value is 100 at `x=\pm 10` and the absolute minimum value is 0 at `x=0`.