Problems from Intervals

Calculate:

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

  1. The center of an interval is: $$$C=\dfrac{a+b}{2}=\dfrac{2-\sqrt{5}}{2}=1-\dfrac{\sqrt{5}}{2}$$$ and the radius is: $$$d(a,C)=d \Big(-\sqrt{5},1-\dfrac{\sqrt{5}}{2}\Big)=\Big|1-\dfrac{\sqrt{5}}{2}+\sqrt{5}\Big|=1+\dfrac{\sqrt{5}}{2}$$$

  2. The lower endpoint is: $$a=C-r=-\dfrac{1}{3}-1=-\dfrac{4}{3},$$ and the uppe endpoint is: $$b=C+r=-\dfrac{1}{3}+1=\dfrac{2}{3}.$$

Solution:

  1. $$C=1-\dfrac{\sqrt{5}}{2}$$ and $$r=1+\dfrac{\sqrt{5}}{2}$$
  2. $$a=-\dfrac{4}{3}$$ and $$b=\dfrac{2}{3}.$$
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Say which of the following affirmations are true or false:

  1. $$\dfrac{1}{\sqrt{5}}$$ belongs to the interval $$\Big[\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{7}}\Big]$$
  2. $$\sqrt{2}$$ belongs to the interval $$\Big[\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{7}}\Big]$$
  3. $$\dfrac{1}{\sqrt{7}}$$ belongs to the interval $$\Big[\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{7}}\Big]$$
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Development:

  1. As $$\sqrt{2} < \sqrt{5} < \sqrt{7}$$, we have that $$\dfrac{1}{\sqrt{5}} \in \Big[\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{7}}\Big]$$

  2. $$\sqrt{2} > \dfrac{1}{\sqrt{7}}$$, and therefore $$\sqrt{2} \notin \Big[\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{7}}\Big]$$

  3. The interval $$\Big[\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{7}}\Big]$$ is closed and bounded, so that the endpoints belong to it.

Solution:

  1. True.
  2. False.
  3. True.
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