Problems from Intervals

Calculate:

The center and the radius of the interval $[- \sqrt{5}, 2] .$
The endpoints of the interval of center $- \frac{1}{3}$ and radius $1$ .

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Development:

The center of an interval is: $C = \frac{a + b}{2} = \frac{2 - \sqrt{5}}{2} = 1 - \frac{\sqrt{5}}{2}$ and the radius is: $d (a, C) = d (- \sqrt{5}, 1 - \frac{\sqrt{5}}{2}) = | 1 - \frac{\sqrt{5}}{2} + \sqrt{5} | = 1 + \frac{\sqrt{5}}{2}$
The lower endpoint is: $a = C - r = - \frac{1}{3} - 1 = - \frac{4}{3},$ and the uppe endpoint is: $b = C + r = - \frac{1}{3} + 1 = \frac{2}{3} .$

Solution:

$C = 1 - \frac{\sqrt{5}}{2}$ and $r = 1 + \frac{\sqrt{5}}{2}$
$a = - \frac{4}{3}$ and $b = \frac{2}{3} .$

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Say which of the following affirmations are true or false:

$\frac{1}{\sqrt{5}}$ belongs to the interval $[\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{7}}]$
$\sqrt{2}$ belongs to the interval $[\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{7}}]$
$\frac{1}{\sqrt{7}}$ belongs to the interval $[\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{7}}]$

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Development:

As $\sqrt{2} < \sqrt{5} < \sqrt{7}$ , we have that $\frac{1}{\sqrt{5}} \in [\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{7}}]$
$\sqrt{2} > \frac{1}{\sqrt{7}}$ , and therefore $\sqrt{2} \notin [\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{7}}]$
The interval $[\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{7}}]$ is closed and bounded, so that the endpoints belong to it.

Solution:

True.
False.
True.

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