I need help in solving (f-g)(5) and finding the domain.

I also need to solve (f/g)(9) and find the domain.

I need help in solving (f-g)(5) and finding the domain.

I also need to solve (f/g)(9) and find the domain.

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Manipulating functions word very similarly to manipulating variables or other numbers. We can add them, subtract, multiply, divide, etc. You just have to keep in mind that the function is a whole "chunk" of information, not just a single number or variable in most cases.

Subtracting two functions:

**f(x) - g(x) is also equal to (f-g)(x)** This literally says to subtract the g function from the f function.

f(x) = 9 + (9/x) AND g(x) = (9/x) SO....

**[9 + (9/x)] - [(9/x)] **Here, you can either plug in the number given in the problem or simplify. If you notice postitive (9/x) and a negative (9/x). They cancel each other out. SO...

**(f-g)(x) = 9 (this is true for all numbers of x, excluding zero)**

(f/g)(x) is the same as f(x)/g(x)

(9 + (9/x)) / (9/x) (since this is a complex fraction, we can flip the denominator and then multiply across)

(9 + 9/x) * (x/9) = 9x/9 + 9x/9x (simplify)

x + 1 (plug in x = 9)

**answer: 9 + 1 = 10**

Concerning domain: there are only 3 instances in which a variable within a function creates a DNE (does not exist) value.

1: 1/x = x cannot equal zero. the denominator can never equal zero, any values for x which turn the denominator to zero is NOT part of the domain of the function.

2: SQRT(neg. number) = evenly rooted numbers cannot be negative, such as the square root of -4 or the fourth root of -16. (these numbers will create imaginary numbers.) any value of x that creates a negative in this situation is not part of the domain.

3: 1/SQRT(neg. number) = this combines the previous two rules, where we cannot have a zero as denominator or a negative number within the even root.

**SO.... for the function f(x), it domain of x can be all real numbers except zero: (-infinity, 0) U (0, +infinity)**

**g(x) domain in all real numbers except zero. the same as f(x)**

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