let f(x)=2/x-1 and g(x)=2x-2/3x+3.find.
a)f+g
b)fg
c)domain of(f+g)and fg
Assuming you mean
f(x)=2/(x-1) and g(x)=(2x-2)/(3x+3)
then g(x) = 2(x-1)/(3(x+1))
f+g = 2/(x-1) + (2/3)(x-1)/(x+1)
= ( 6(x+1) + 2(x-1)(x-1))/(3(x^2 - 1))
= (6x + 6 + 2x^2 - 4x + 2)/(3(x^2 - 1)
= (2/3)(x^2 + x + 4)/(x^2 - 1)
you do fg, which means you just multiply them
domain of f+g is x∈R, x ≠ ±1
HELLO MS. SUE THIS PERSON NEEDS HELP
mesfinabrham9
Domain of(f+g)and fg
a) To find f+g, we will simply add the functions f(x) and g(x):
f(x) = 2/(x-1)
g(x) = (2x-2)/(3x+3)
f+g = 2/(x-1) + (2x-2)/(3x+3)
b) To find fg, we will multiply the functions f(x) and g(x):
fg = (2/(x-1))[(2x-2)/(3x+3)]
c) To find the domain of f+g and fg, we need to consider any values of x that would cause division by zero. In this case, we need to check if the denominators (x-1) and (3x+3) are equal to zero.
For f+g:
x-1 = 0
x = 1
So the domain of f+g is all real numbers except x=1.
For fg:
x-1 = 0
x = 1
3x+3 = 0
3x = -3
x = -1
So the domain of fg is all real numbers except x=1 and x=-1.
To find the answers to these questions, we need to understand how to perform arithmetic operations on functions and determine the domain of resulting functions.
a) To find f+g, we add the two functions together. Since f(x) = 2/(x-1) and g(x) = (2x-2)/(3x+3), we can write f+g as follows:
f+g = 2/(x-1) + (2x-2)/(3x+3)
To simplify this expression, we need to find the common denominator of the two fractions:
The common denominator is (x-1)(3x+3). We can rewrite the fractions with this common denominator:
f+g = (2*(3x+3))/((x-1)(3x+3)) + ((2x-2)*(x-1))/((x-1)(3x+3))
Now, we can combine the fractions:
f+g = (6x + 6 + 2x^2 - 2x - 2)/((x-1)(3x+3))
Simplifying further, we have:
f+g = (2x^2 + 4x + 4)/((x-1)(3x+3))
b) To find fg, we multiply the two functions together:
fg = (2/(x-1)) * ((2x-2)/(3x+3))
Multiplying the numerators and denominators gives:
fg = (4x-4)/(3x^2-x-3)
c) To find the domain of f+g, we need to consider any restrictions on x that would make the function undefined. In this case, the function f+g will be undefined if the denominator (x-1)(3x+3) is equal to zero. Therefore, we need to find the values of x that make either (x-1) or (3x+3) equal to zero.
Setting x-1=0 gives x=1, and setting 3x+3=0 gives x=-1. Therefore, the domain of f+g is all real numbers except x=1 and x=-1.
To find the domain of fg, we also need to consider any restrictions on x that would make the function undefined. In this case, the function fg will be undefined if the denominator (3x^2-x-3) is equal to zero. Therefore, we need to find the values of x that make (3x^2-x-3) equal to zero.
Using the quadratic formula, we can solve for x:
x = (-(-1) ± √((-1)^2 - 4(3)(-3))) / (2(3))
Simplifying, we have:
x = (1 ± √(1 + 36)) / 6
Simplifying further, we have:
x = (1 ± √37) / 6
Therefore, the domain of fg is all real numbers except x = (1 ± √37) / 6.