# Pre cal

1.Find the domain of the function algebraically
y=3x-1/(x+3)(x-1)

y=(1/x)+5/(x-3)

2.Find the range of the function
y=10-x^2

y=(3+x^2)/(4-x^2)

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1. all x except x=-3 or x=1

all x except x=0 or x=3

2. range depends on the domain. if all x is allowed, then rande is all real numbers

if domain is all real numbers, then range is -1to inf (think, what if x=2, what if x= inf, what if x=-inf)

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2. domain is your choice of x's that will give you a valid y
Remember we can't divide by zero
so (x+3)(x-1) ≠ 0
so when do we get a zero ?
That happens when x = -3 or x=1

so Domain: all values of x except x = -3 or x = 1

2 .
The range of a function is the resulting values of y which are valid
y = 10 - x^2
this is a parabola which opens downwards and the vertex is (0,10)
so the range is : y ≤ 10

3. Perhaps looking at the graph will let you decide what the range is
http://www.wolframalpha.com/input/?i=range+of+y%3D%283%2Bx%5E2%29%2F%284-x%5E2%29+
notice that any ±x will yield the same y
so there is symmetry about the x axis
when x = 0 , y = 3/4
as x ---> ∞ , y ---> -1

rangle: y ≥ 3/4 , y < -1

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posted by Reiny

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