Question #04a42

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Question #04a42

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Answer 1 · 2017-11-25T17:09:45

$2 \geq y \geq 0$ $2>=y>=0$

Explanation:

The range of a function is the range of values which the function can take given it's domain.
As the domain is not given, assume that it is all values of x which give a real y (in this case $| x | \leq 2$ $|\x|<=2$ ).

Three important facts to remember in this question are that:
a) you can only take the square root of a positive number
b) $y = \sqrt{f (x)}$ $y=sqrt(f(x))$ only plots the positive square root (principal root)
c) any real number squared $\geq 0$ $>=0$

Using a), you know $4 - x^{2} \geq 0$ $4-x^2>=0$ . Because of c), the maximum value $4 - x^{2}$ $4-x^2$ can take is 4, when $x = 0$ $x=0$ , so the maximum value of y is $\sqrt{4} = 2$ $sqrt(4)=2$ .
Using a) the minimum value $4 - x^{2}$ $4-x^2$ can take is 0, when $x = \pm 2$ $x=+-2$ , and $\sqrt{0} = 0$ $sqrt(0)=0$ .

Given the minimum and maximum it follows that $0 \leq x \leq 2$ $0<=x<=2$

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Question #04a42

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