I need to find the domain of f, the excluded values of f, the domain of (fg)(x); find (f+g)(-5), (f-g)(x), and (fg)(x) to the problem of:
Let f(x)=(7x+1)/(3x+2) and g(x)=6x+5.
excluded values for the domain of f: when 3x+2 = 0
Otherwise the domain of f and g is all reals.
f+/-g = (x+1)/(3x+2) +/- (6x+5), so same domain as f.
fg = (7x+1)(6x+5)/(3x+2), so the domain is the same as that of f
(f+g)(-5) = f(-5)+g(-5) = -34/-13 - 25
To find the domain of a function, you need to determine the set of all possible input values for which the function is defined.
1. Domain of f(x):
In this case, f(x) is a rational function. For a rational function, the domain will exclude any values of x that would result in division by zero. So, to find the domain of f(x), we need to determine the values of x where the denominator of f(x) is equal to zero.
The denominator of f(x) is 3x + 2. To find the values of x that make the denominator zero, we set 3x + 2 = 0 and solve for x:
3x + 2 = 0
3x = -2
x = -2/3
Therefore, the value x = -2/3 is the only value that would make the denominator zero. So, the domain of f(x) will exclude x = -2/3. Therefore, the domain of f(x) is all real numbers except x = -2/3.
2. Excluded values of f(x):
As mentioned above, the value x = -2/3 is excluded from the domain of f(x) because it would result in division by zero. So, x = -2/3 is the excluded value for f(x).
3. Domain of (fg)(x):
To find the domain of the composition of two functions, (fg)(x), we need to find the values of x that make both f(x) and g(x) defined.
Since the domain of f(x) is all real numbers except x = -2/3, and the domain of g(x) is all real numbers, the domain of (fg)(x) will be the intersection of the domains of f(x) and g(x).
Since there are no restrictions on the domain of g(x), the only restriction comes from the domain of f(x). Therefore, the domain of (fg)(x) will be all real numbers except x = -2/3.
4. (f+g)(-5):
To find (f+g)(-5), you substitute -5 for x in both f(x) and g(x), and then add the two results.
First, evaluate f(x) at x = -5:
f(x) = (7x + 1)/(3x + 2)
f(-5) = (7(-5) + 1)/(3(-5) + 2)
f(-5) = (-34)/(-13)
f(-5) = 34/13
Then, evaluate g(x) at x = -5:
g(x) = 6x + 5
g(-5) = 6(-5) + 5
g(-5) = -25
Finally, add the two results:
(f+g)(-5) = (34/13) + (-25)
(f+g)(-5) = 34/13 - 25/1 (common denominator is 13)
(f+g)(-5) = (34 - 325)/13
(f+g)(-5) = -291/13
Therefore, (f+g)(-5) = -291/13.
5. (f-g)(x):
To find (f-g)(x), you substitute x for both f(x) and g(x), and then subtract the two results.
First, evaluate f(x):
f(x) = (7x + 1)/(3x + 2)
Then, evaluate g(x):
g(x) = 6x + 5
Finally, subtract the two:
(f-g)(x) = (7x + 1)/(3x + 2) - (6x + 5)
(f-g)(x) = (7x + 1)/(3x + 2) - (6x + 5)(3x + 2)/(3x + 2) (to get a common denominator)
(f-g)(x) = (7x + 1 - (18x + 10))/(3x + 2)
(f-g)(x) = (7x - 18x + 1 - 10)/(3x + 2)
(f-g)(x) = (-11x - 9)/(3x + 2)
Therefore, (f-g)(x) = (-11x - 9)/(3x + 2).
6. (fg)(x):
To find (fg)(x), you multiply f(x) and g(x).
First, evaluate f(x):
f(x) = (7x + 1)/(3x + 2)
Then, evaluate g(x):
g(x) = 6x + 5
Finally, multiply the two:
(fg)(x) = (7x + 1)/(3x + 2) * (6x + 5)
(fg)(x) = (42x^2 + 35x + 6x + 5)/(3x + 2)
(fg)(x) = (42x^2 + 41x + 5)/(3x + 2)
Therefore, (fg)(x) = (42x^2 + 41x + 5)/(3x + 2).