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

Jul 27, 2015

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

#### Explanation:

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

You can also use calculus (just a little preview if you are in precalculus). Since the derivative $f ' \left(x\right) = 2 x$, 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 $x = \setminus \pm 10$ of the closed and bounded interval $\left[- 10 , 10\right]$:

$f \left(0\right) = {0}^{2} = 0$, $f \left(\setminus \pm 10\right) = {10}^{2} = 100$.

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