# precalc

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find the domain and range of the following as an inequality and interval notation.

y=sqrtx-3

2: y=3x+5

3: y= 2/x-7

4; y=2+sqrt x

• precalc -

domain = 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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