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How do you find domain and range for #y=sqrt((4 - x²))#?

Answer 1

Isaiah Anthony

Refer to explanation

We have that because it's a square root.

#4-x^2>=0=>(2-x)*(2+x)>=0# which holds for #[-2,2]# hence

#D(f)=[-2,2]#

The scope is

#y=sqrt(4-x^2)=>y^2=4-x^2=>x^2=4-y^2>=0=>(2-y)(2+y)>=0#

Because #y>=0# we have that #R(f)=(0,2)#

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Answer 2

Savannah Anderson

To find the domain of the function ( y = \sqrt{4 - x^2} ), we need to determine the values of ( x ) for which the expression under the square root is non-negative. Since the square root of a negative number is not real, ( 4 - x^2 ) must be greater than or equal to zero:

[ 4 - x^2 \geq 0 ]

Solving this inequality:

[ 4 \geq x^2 ]

[ -4 \leq x^2 ]

[ -2 \leq x \leq 2 ]

So, the domain of the function is ( -2 \leq x \leq 2 ).

To find the range, notice that the square root of any non-negative number is always non-negative. Therefore, the range of the function ( y = \sqrt{4 - x^2} ) is all non-negative real numbers, or ( y \geq 0 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.