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)

Please help me understand how to do them.

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)

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

To find the domain of a function algebraically, you need to identify any values of x that would lead to the function being undefined. Here's how to do it:

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

To determine the domain, we need to look for any values of x that would make the denominator equal to zero. In this case, the denominator is (x+3)(x-1).

For the denominator (x+3)(x-1) to be zero, either (x+3) must be zero or (x-1) must be zero.

(x+3) = 0 --> x = -3
(x-1) = 0 --> x = 1

Therefore, the denominator is zero when x = -3 or x = 1. These are the values where the function is undefined. So, the domain of the function is all real numbers except -3 and 1. In interval notation, the domain is (-∞, -3) U (-3, 1) U (1, +∞).

2. Find the domain of the function y = (1/x) + (5/(x-3)):

For this function, we only need to consider the first term (1/x) and the second term (5/(x-3)) separately since the denominator of each term must not be zero.

The first term is (1/x), which means x cannot be zero. So, x ≠ 0.

The second term is (5/(x-3)), which means (x-3) cannot be zero. So, x ≠ 3.

Therefore, the domain of the function is all real numbers except 0 and 3. In interval notation, the domain is (-∞, 0) U (0, 3) U (3, +∞).

Now let's move on to finding the range of the given functions:

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

To determine the range, we need to look at the possible values of y as we vary x. The highest value of x^2 is positive infinity (as x approaches infinity), and the lowest value of x^2 is zero (when x = 0). So, the highest value of 10 - x^2 is 10 and the lowest value is 10 - 0 = 10. Therefore, the range of this function is (-∞, 10].

2. Find the range of the function y = (3 + x^2)/(4 - x^2):

To find the range, we should consider the values x^2 can take. Since x^2 is always non-negative, it can range from 0 to positive infinity. However, we also need to consider the denominator (4 - x^2). If this denominator equals zero, the function becomes undefined.

To determine when (4 - x^2) = 0, we solve the equation:
4 - x^2 = 0
x^2 = 4
x = ±2

So, the function is undefined when x = ±2. These are the values where the denominator becomes zero.

Since x^2 can vary from 0 to positive infinity, except for x = ±2, the range of the function is (-∞, +∞) excluding the values the function is undefined at (y = 3 + x^2)/(4 - x^2) when x = ±2.