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

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

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Answer 1 · 2015-07-27T18:46:11

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

Explanation:

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

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 $x=0$ .

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

1 Answer

Explanation:

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Over the x-value interval [−10,10][−10,10][−10,10], what are the absolute extrema of f(x)=x^2f(x)=x2f(x)=x^2?

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Explanation:

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