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

1 Answer
Jul 27, 2015

Answer:

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#.