[Note: I really need help on these Domain questions. I've been struggling with this all day]
-Consider the following:
f(x)= 5/x , g(x)= 7/(x+7)
-Find the solutions to the following:
1). (f+g)(x)
2). Domain of (f+g)(x)
[Use interval notation too]
3). (f-g)(x)
4). Domain of (f-g)(x)
[Use interval notation too]
5). (fg)(x)
6). Domain of (fg)(x)
[Use interval notation too]
7). (f/g)(x)
8). Domain of (f/g)(x)
[Use interval notation too]
[Extra Note: You don't have to do interval notation. Just relaying information]
The denominator cannot = 0
For f(x) x cannot = 0
for g(x) x cannot = -7
1) If you add the two fractions together, the denominator will be
x(x+7)
Domain would be from
(-infinity, -7)U(-7,0)U(0,infinity)
Using ( ) because those values aren't included in the domain.
3) you are subtracting so the denominator will be the same as above and your domain will be the same.
Can you finish from here?
the domains include all real numbers except where the denominator is zero.
So, figure the resultant function, and then exclude any places where the denominator is zero.
AND, exclude 0 and -7 because at those values f or g is undefined, so even though some mixture of them might look ok there, f and g have to be evaluated to come up with the result.
So, whaddya get?
Sure, I'd be happy to help you with these domain questions!
1). To find (f+g)(x), you need to add the two functions f(x) and g(x) together:
(f+g)(x) = f(x) + g(x)
= 5/x + 7/(x+7)
2). To determine the domain of (f+g)(x), you need to identify the values of x that make the function undefined. This occurs when the denominator is equal to zero because division by zero is undefined. In this case, you need to find the values of x that make either 5/x or 7/(x+7) undefined.
For 5/x, x cannot be equal to zero because dividing by zero is undefined. Similarly, for 7/(x+7), x cannot be equal to -7 because it would result in division by zero. Therefore, the domain of (f+g)(x) is all real numbers except x = 0 and x = -7.
In interval notation, you can express the domain as (-∞, -7) U (-7, 0) U (0, ∞).
3). To find (f-g)(x), subtract g(x) from f(x):
(f-g)(x) = f(x) - g(x)
= 5/x - 7/(x+7)
4). Similar to question 2, you need to identify the values of x that make (f-g)(x) undefined. Again, this occurs when the denominator is equal to zero.
For 5/x, x cannot be equal to zero. For 7/(x+7), x cannot be equal to -7. So, the domain of (f-g)(x) is the same as that of (f+g)(x): all real numbers except x = 0 and x = -7.
The domain of (f-g)(x) in interval notation is (-∞, -7) U (-7, 0) U (0, ∞).
5). To find (fg)(x), you need to multiply f(x) by g(x):
(fg)(x) = f(x) * g(x)
= (5/x) * (7/(x+7))
6). Regarding the domain of (fg)(x), you need to consider the values of x that make the function undefined. This occurs when either 5/x or 7/(x+7) is undefined.
For 5/x, x cannot be equal to zero. For 7/(x+7), x cannot be equal to -7. Thus, the domain of (fg)(x) is the same as that of (f+g)(x) and (f-g)(x): all real numbers except x = 0 and x = -7.
The domain of (fg)(x) in interval notation is (-∞, -7) U (-7, 0) U (0, ∞).
7). To find (f/g)(x), divide f(x) by g(x):
(f/g)(x) = f(x) / g(x)
= (5/x) / (7/(x+7))
8). Determining the domain of (f/g)(x), you need to identify the values of x that make the function undefined. Division by zero occurs when the denominator is equal to zero.
For 5/x, x cannot be equal to zero. For 7/(x+7), x cannot be equal to -7. Therefore, the domain of (f/g)(x) is the same as the previous functions: all real numbers except x = 0 and x = -7.
The domain of (f/g)(x) in interval notation is (-∞, -7) U (-7, 0) U (0, ∞).
I hope this explanation helps you understand how to find the solutions and domains for these functions. Let me know if you have any further questions!