1.for the given function f&g, find the following and state the domain of each result.


f(x)=2x+1/9x-5; g(x) 4x/9x-5
a.(f+g)(x)=?
What is the domain?

b. (f/g)(x)=?
what is the domain?

2.the function f is defined as follows
{x+6 if -5 ¡Üx<1
f(x)={9 if x=1
{-x+4 if x>1

a. the domain of the function f is ?
b. what are the intercepts?

on the second question it should be

f(x)= {x+6 if -5is less than or greater than x<1
9 if x=1
-x+4 if x>1

sorry it didn't come out right.

it is customary online to use <= or >= for less than or equal, etc.

for both f and g, the domain is all reals except x = 5/9
so, f+g has the same domain

f/g = (2x+1)/4x
so, you would think the domain is all reals except x=0, but since neither f nor g is defined at x = 5/9, we have to exclude that as well.

2.
since all pieces of f are polynomials, their domain is just as stated: [-5,1)U(1)U(1,oo) = [-5,oo)
intercepts: (0,6) and (4,0)

To find the sum and quotient of functions, we need to add or divide the corresponding terms of the functions. Let's calculate each part step by step.

1a. (f+g)(x) = f(x) + g(x)
To find the sum of f(x) and g(x), we add the corresponding terms:
(f+g)(x) = (2x + 1) / (9x - 5) + (4x) / (9x - 5)

To add the fractions, we need a common denominator, which is already (9x - 5) in this case. Therefore:

(f+g)(x) = (2x + 1 + 4x) / (9x - 5)
= (6x + 1) / (9x - 5)

The domain is determined by the values that x cannot take, as they may result in undefined values. In this case, x cannot be equal to 5/9 since it would make the denominator zero, leading to division by zero. Therefore, the domain of (f+g)(x) is all real numbers except x = 5/9.

1b. (f/g)(x) = f(x) / g(x)
To find the quotient of f(x) and g(x), we divide the corresponding terms:
(f/g)(x) = (2x + 1) / (9x - 5) / (4x) / (9x - 5)

To divide fractions, we invert the second fraction and multiply:
(f/g)(x) = (2x + 1) / (9x - 5) * (9x - 5) / (4x)

Simplifying, we have:
(f/g)(x) = (2x + 1) / (4x)

The domain is determined by the values that x cannot take, as they may result in undefined values. In this case, x cannot be equal to 0 since it would make the denominator zero, leading to division by zero. Therefore, the domain of (f/g)(x) is all real numbers except x = 0.

2a. The domain of the function f is determined by the values of x for which each piece of the function is defined.
For -5 ≤ x < 1, the function is defined as: f(x) = x + 6
For x = 1, the function is defined as: f(x) = 9
For x > 1, the function is defined as: f(x) = -x + 4

Thus, the domain of the function f is -5 ≤ x < 1 and x > 1.

2b. To find the intercepts, we need to consider when the function f(x) crosses the x-axis (y = 0) and the y-axis (x = 0).

For y-intercept (x = 0), we substitute x = 0 into each piece of the function:

For -5 ≤ x < 1: f(0) = 0 + 6 = 6
For x = 1: f(0) = 9
For x > 1: f(0) = -(0) + 4 = 4

Therefore, the y-intercept is (0, 6) when -5 ≤ x < 1 and x > 1, and (0, 9) for x = 1.

For x-intercept (y = 0), we solve each piece of the function for x = 0:

For -5 ≤ x < 1: x + 6 = 0 => x = -6
For x > 1: -x + 4 = 0 => x = 4

Therefore, the x-intercepts are (-6, 0) when -5 ≤ x < 1, and (4, 0) for x > 1.

To find the results of the given functions and determine their respective domains, we will need to substitute the given expressions into the given equations. Let's solve each question step by step:

1. a. To find (f+g)(x), we add the functions f(x) and g(x):

f(x) = (2x+1)/(9x-5)
g(x) = (4x)/(9x-5)

(f+g)(x) = f(x) + g(x)
= (2x+1)/(9x-5) + (4x)/(9x-5)
= (2x+1+4x)/(9x-5)
= (6x+1)/(9x-5)

The domain of (f+g)(x) will be the same as the domain of f(x) and g(x), which is all real numbers except x = 5/9 (since it would result in a division by zero).

Therefore, the domain of (f+g)(x) is (-∞, 5/9) U (5/9, ∞).

b. To find (f/g)(x), we divide the function f(x) by g(x):

f(x) = (2x+1)/(9x-5)
g(x) = (4x)/(9x-5)

(f/g)(x) = f(x) / g(x)
= [(2x+1)/(9x-5)] / [(4x)/(9x-5)]
= (2x+1)/(4x)

The domain of (f/g)(x) will exclude any values of x that make the denominator zero, which is x = 0.

Therefore, the domain of (f/g)(x) is (-∞, 0) U (0, ∞).

2. a. To find the domain of the function f, we need to consider the different cases defined in the function:

For -5 ≤ x < 1: The function is defined as f(x) = x+6, which is valid for all real numbers within the given range.

For x = 1: The function is defined as f(x) = 9.

For x > 1: The function is defined as f(x) = -x+4, which is valid for all real numbers greater than 1.

Therefore, the domain of f is (-∞, 1) U (1, ∞).

b. To find the intercepts of the function f, we can set f(x) = 0 and solve for x:

Case 1: x+6 = 0
x = -6

Case 2: -x+4 = 0
x = 4

Therefore, the x-intercepts (where the function crosses the x-axis) are x = -6 and x = 4. There are no y-intercepts (where the function crosses the y-axis) since it would require f(0), but f(0) is not defined in the given function.