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Mathematics 12 Online
OpenStudy (anonymous):

(f/g) (x)=? I know the answer is (8x+32)/(3x-21). but what is the domain? isnt it (-infinity , 7) U (7 , +infinity)? I wrote that but it is wrong...

OpenStudy (ranga):

The domain looks right to me.

OpenStudy (anonymous):

but it is wrong... I dont know why......

OpenStudy (ranga):

Is this an online question or a textbook question?

OpenStudy (anonymous):

Yea it is a online question

OpenStudy (ranga):

Try putting infinity instead of +infinity.

OpenStudy (anonymous):

Same...

OpenStudy (ranga):

Sometimes online software gets thrown off if the result is not exactly as it expects it to be. Get rid of the extra space before the comma in (-infinity ,7) and in (7 ,infinity).

OpenStudy (anonymous):

um.... Yea.. I have try that for several times like (-infinity,7)U(7,infinity),(-infinity, 7) and in (7 ,infinity)...etc but they r all incorrect... maybe i should go asking my professor...

OpenStudy (ranga):

But in the last one I still see the extra space in the third line from the end.

OpenStudy (anonymous):

Yea i mean i try several ways.... it said "You have attempted this problem 24 times. " lol

OpenStudy (ranga):

But you have the correct answer right from the first attempt. You just may have to ask your teacher why it keeps saying it is wrong. The only x value that is not allowed is x = 7. So the domain is all values of x except 7 and that is exactly what you have stated.

OpenStudy (ranga):

If you find out the answer do post it. I am curious to know what your teacher has to say.

OpenStudy (anonymous):

OK haha thx anyway

OpenStudy (ranga):

you are welcome.

OpenStudy (anonymous):

he said, "You have to find for what value x does 3/(x+4)=7 That way, the denominator you initially started with when you first plugged g in f is zero"

OpenStudy (ranga):

But that was NOT the question posted here. The question was: What is the domain of (8x+32) / (3x-21) ?

OpenStudy (anonymous):

Yea.. the full question is " Given that f(x)=[ 8/(x−7)] and g(x)=[ 3/(x + 4)], find (d) (f/g) (x)=____, and its domain is_____"

OpenStudy (anonymous):

so maybe we have to use that equation?

OpenStudy (ranga):

Oh, OK. The first thing we notice is g(x) is undefined when x = -4 and f(x) is undefined when x = 7 Also when finding f/g you are dividing something by g. Whenever dividing anything you will have to first make sure you are NOT dividing by zero.

OpenStudy (anonymous):

so.. i solve that x = -25/7...

OpenStudy (ranga):

How did you get that?

OpenStudy (anonymous):

He said I have to find for what value x does 3/(x+4)=7 so 3=7(x+4)=7x+28 so x is equal to -25/7?

OpenStudy (anonymous):

omg my professor just said "Whoops disregard that message. I though it was f of g of x." "But it's similar. So you have to start with putting f/g. Before simplifying, what values of x male zeros on the bottom?"

OpenStudy (ranga):

But you would do that if we have to find f(g(x)). f(x) = 8/(x - 7) and x cannot be 7 So to find f(g(x)) we have to make sure g(x) is not 7 So for what value does g(x) = 7? At x = -25/7 But for dividing f/g(x) I am not sure why we need to do that.

OpenStudy (anonymous):

um...

OpenStudy (anonymous):

OH!!!! I FINALLY FOUND THE ANSWER!!! Its (-infinity , -4)U(-4, 7)U(7, infinity).

OpenStudy (ranga):

Exactly! Look at my answer about 8 replies before. f(x) is not defined for x = -4 and g(x) is not defined for x = 7 So (-infinity, -4) U (-4, 7) U (7, infinity) If the original f(x) and g(x) were given in the problem this is what we would have arrived at.

OpenStudy (ranga):

switch f and g in line 2 above.

OpenStudy (anonymous):

Yea.. I finally know we should find the domain before simplified.

OpenStudy (ranga):

Exactly! That is the moral of the story.

OpenStudy (anonymous):

haha thx!!!

OpenStudy (ranga):

BTW, the answer is good for all operations involving f and g. Addition, subtraction, multiplication, division, etc. Because for one function x = -4 is not allowed and for the other x = 7 allowed. So these two numbers have to be excluded when these two functions are combined in any fashion.

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