College Algebra - Functions f(x)/g(x)

[protectordelafe]

Original Values:

1. Solve:


2. Find the domain of

My Solution was {x| x≠ 0}
Correct Answer: {x| x≠ 0, x≠7/9} (According to the testing software)

3. I don't understand why I got this wrong
The correct solution contains x ≠ 7/9. How was this extra value calculated?
From what I've learned so far, for this particular problem, all you have to do is set the denominator equal to zero and solve)
The denominator only contains a 9x, which would be 9x=0, and once solved would be x=0.
Thanks in advance!

What I've learned so far -
-Denominator must not equal zero to avoid "Undefined" Cannot divide by Zero.
-Square root should be greater or equal to 0 in the numerator and greater than 0 in the Denominator

 

[MarkFL]

Hello, and welcome to FMH!

You must also consider the domains of the original functions, both of which are undefined for \(x=\dfrac{7}{9}\).

 

[protectordelafe]

Thanks! I was asked to add {F(x); G(x)}, Multiply, subtract and I did get those right. But the division didn't seem to make sense. Is that always the case when doing fractions, to expect 2 answers at least?

 

[HallsofIvy]

Thanks! I was asked to add {F(x); G(x)}, Multiply, subtract and I did get those right. But the division didn't seem to make sense. Is that always the case when doing fractions, to expect 2 answers at least?

Two answers to what question?

 

[protectordelafe]

Two answers to what question?

Two answers when asked to find the domain of two functions being divided, as seen in the correct solution under #2. Find the Domain. This question was in response to the answer given by MarkFL

 

[tkhunny]

Ponder with me, please. How are these two functions different?

[math]h(x) = \dfrac{(x-7)\cdot x}{x}\;and\;j(x) = x-7[/math]

 

[Deleted member 4993]

Original Values: View attachment 16479

1. Solve:
View attachment 16481

2. Find the domain of View attachment 16480

My Solution was {x| x≠ 0}
Correct Answer: {x| x≠ 0, x≠7/9} (According to the testing software)

3. I don't understand why I got this wrong
The correct solution contains x ≠ 7/9. How was this extra value calculated?
From what I've learned so far, for this particular problem, all you have to do is set the denominator equal to zero and solve)

You factored out or divided by (9x -7), while evaluating [f/g]

For that to be valid, the implied assumption is x ≠ 7/9 [because at x = 7/9, we would have (9x -7) = 0 and to cancel out (9x -7) you have to divide by (9x -7)].

The denominator only contains a 9x, which would be 9x=0, and once solved would be x=0.
Thanks in advance!

What I've learned so far -
-Denominator must not equal zero to avoid "Undefined" Cannot divide by Zero.
-Square root should be greater or equal to 0 in the numerator and greater than 0 in the Denominator

 

[JeffM]

You are greatly complicating something simple. Domains are case by case.

g(0) = 0 so you cannot have [MATH]\dfrac{f(0)}{g(0)}.[/MATH] Obvious.

g(x) and f(x) are not defined at x = 7/9 so you cannot have [MATH]\dfrac{f(7/9)}{g(7/9)}.[/MATH]
What is confusing you is this

[MATH] \text {IF } x \ne 7/9 \text { and } x \ne 0, \text { THEN } \dfrac{f(x)}{g(x)} \text { is defined and}[/MATH]
[MATH]\dfrac{f(x)}{g(x)} = \dfrac{7x + 9}{9x - 7} \div \dfrac{9x}{9x - 7} = \dfrac{7x + 9}{\cancel {9x - 7}} * \dfrac{\cancel {9x - 7}}{9x} = \dfrac{7x + 9}{9x}.[/MATH]
When you look just at (7x + 9)/(9x), it does exist at 7/9. But that is irrelevant because

[MATH]\text {IF } x = \dfrac{7}{9},\text { THEN } \dfrac{f(x)}{g(x)} \ne \dfrac{7x + 9}{9x}[/MATH]
because f(x)/g(x) is not even defined at x = 7/9.

 

[protectordelafe]

Thanks, everyone!!!