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December 20, 2014

December 20, 2014

Posted by **anon** on Wednesday, May 4, 2011 at 9:36pm.

y=sqrtx-3

2: y=3x+5

3: y= 2/x-7

4; y=2+sqrt x

- precalc -
**Jai**, Wednesday, May 4, 2011 at 9:53pmdomain = set of all possible values of x

range = set of all possible values of y

therefore,

(1) y = sqrt(x-3)

note that the term inside the sqrt can be zero but cannot be negative, for it will become imaginary. thus,

Domain: x must be greater than or equal to 3.

also, y can be zero (if x=3) but can never be negative since sqrt of any positive real number is always positive, thus,

Range: y must be greater than or equal to 0.

(2) y = 3x + 5

this is an equation of a line, and note that there is neither restriction in both x and y, thus,

Domain: x can be all real numbers

Range: y can be all real numbers

(3) y = 2/(x-7)

note that denominators cannot be 0, thus,

Domain: x can be all real numbers except 7.

for the range, observe that y cannot be equal to zero, because, if so,

0 = 2/(x-7) , then cross-multiplying,

0 = 2 , which is wrong, thus

Range: y is all real numbers except 0

(4) y = 2 + sqrt(x)

note that this is almost similar to #1, except that there is +2. therefore,

Domain: x must be greater than or equal to zero.

for the range, the minimum possible value of x is zero, but because there is +2,

Range: y must be greater than or equal to 2.

hope this helps~ :D

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