Given f(x)=5x/x-4 and g(x)=6/x+2 find the following
a) (f+g)(x)
b) the domain of (f+g)(x) in interval notation
c) (f-g)(x)
d) the domain of (f-g)(x) in interval notion
e) (f*g)(x)
f) the domain of (f*g)(x)
g) (f/g)(x)
h. the domain of (f/g)(x) in interval notation
Any ideas?
The fractions are throwing me off plus there is an x on top of one of them.
Really? By precalc, fractions should not be troubling you. And don't be afraid of using parentheses to make sure your expressions are correct.
(f+g)(x) = f(x) + g(x) = 5x/(x-4) + 6/(x+2) = (5x^2+16x-24) / ((x+2)(x-4))
since division by zero is undefined, the domain is (-∞,-2)U(4,∞)
Now try the others. Since the original domains exclude -2 and 4, the composite domain will also, even if its simplified form allows them.
To find the answers, we need to perform the operations (+, -, *, /) on the given functions f(x) and g(x). Let's go through each question step by step:
a) (f+g)(x):
To find (f+g)(x), we need to add the functions f(x) and g(x). So (f+g)(x) = f(x) + g(x).
Substituting the given functions into this equation, we have:
(f+g)(x) = (5x/(x-4)) + (6/(x+2))
b) The domain of (f+g)(x) in interval notation:
To determine the domain of a function, we need to identify any values of x that would make the function undefined. In this case, we need to look out for any division by zero situations.
For (f+g)(x), the only function with a potential division by zero is g(x) with the term (x+2) in the denominator.
To find the domain, we set the denominator of g(x) equal to zero and solve for x:
x + 2 = 0
x = -2
Therefore, the domain of (f+g)(x) is all real numbers except x = -2. In interval notation, it can be written as: (-∞, -2) ∪ (-2, ∞).
c) (f-g)(x):
To find (f-g)(x), we need to subtract the functions g(x) from f(x). So (f-g)(x) = f(x) - g(x).
Substituting the given functions into this equation, we have:
(f-g)(x) = (5x/(x-4)) - (6/(x+2))
d) The domain of (f-g)(x) in interval notation:
Similar to the previous case, we need to identify any x-values that would make the function undefined. The only function with a potential division by zero situation is g(x) with the term (x+2) in the denominator.
Setting the denominator of g(x) equal to zero and solving for x:
x + 2 = 0
x = -2
So, the domain of (f-g)(x) is all real numbers except x = -2. In interval notation, it can be written as: (-∞, -2) ∪ (-2, ∞).
e) (f*g)(x):
To find (f*g)(x), we need to multiply the functions f(x) and g(x). So (f*g)(x) = f(x) * g(x).
Substituting the given functions into this equation, we have:
(f*g)(x) = (5x/(x-4)) * (6/(x+2))
f) The domain of (f*g)(x) in interval notation:
Again, we need to identify any x-values that would make the function undefined. We see that both f(x) and g(x) have denominators, so we need to ensure that neither denominator equals zero.
Setting the denominators of f(x) and g(x) equal to zero and solving for x:
For f(x):
x - 4 = 0
x = 4
For g(x):
x + 2 = 0
x = -2
Therefore, the domain of (f*g)(x) is all real numbers except x = 4 and x = -2. In interval notation, it can be written as: (-∞, -2) ∪ (-2, 4) ∪ (4, ∞).
g) (f/g)(x):
To find (f/g)(x), we need to divide the function f(x) by g(x). So (f/g)(x) = f(x) / g(x).
Substituting the given functions into this equation, we have:
(f/g)(x) = (5x/(x-4)) / (6/(x+2))
h) The domain of (f/g)(x) in interval notation:
Similar to the previous cases, we need to identify any x-values that would make the function undefined. For the function (f/g)(x), the denominator of g(x) should not equal zero.
Setting the denominator of g(x) equal to zero and solving for x:
x + 2 = 0
x = -2
So, the domain of (f/g)(x) is all real numbers except x = -2. In interval notation, it can be written as: (-∞, -2) ∪ (-2, ∞).