Stuart R. answered • 08/02/13
Math Tutor - Online and In Home -20 Years Experience
if I understand , the question is: x divided by y = x+9. is it a function? if not find a value that makes it "not a function.
Going on this assumption....
step 1) x/y = x+9. solve for y. y=x/(x+9). therefore f(x) = x /(x+9)
step 2) look at the right hand side of the equation! x+9 will ALWAYS be bigger than x !!
step 3) if it is not a function then there is at least ONE point for x where there is MORE than one y.
step 4)BUT if x is ALWAYS < x+9 , step 3 will never happen.
therefore f(x) which is the same as y IS a function.
the domain of f(x) is all the real numbers that x can be and still give a value for y which is f(x) (same thing).
step 5) notice that when x = -9 we get 9 / (-9+9) which = 9 / 0, which has NO solution (can't divide by zero!).
so the domain of this function is all real numbers other than -9.
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If your problem should read x|y| = x+9 then........
|y| = (x+9)/x
so.... x can NEVER be Zero therefore 0 is NOT in the domain.
the domain is all real numbers other than 0.
By taking the absolute value of all negative outputs (y's), you change them to positive. IN OTHER WORDS (x,y) becomes (x,|y|).
Oh... it's still a function.
I tutor on-line and in home.
Hassan H.
In the case that the equation reads
x|y| = x+9,
Stuart has correctly solved for the explicit relation for y in terms of x:
|y| = (x+9)/x.
However, the latter is not the definition of a function in the independent variable x, since, for instance, letting x=1 produces
|y| = 10,
which implies both y = 10 and y = -10. Therefore, y is not a function of x.
Regards,
Hassan H.
08/02/13
Stuart R.
Hassan;
Of course, you are right.
Stuart
08/02/13
Stuart R.
Oh, by the way, Hassan, which one value of x would yield both 10 and - 10?
08/04/13
Hassan H.
Nice explanation, Stuart! It just occurred to me looking at this question that maybe the questioner means to write absolute values around the y---making the equation into
x|y| = x+9.
Maybe you could cover that possibility, too, since then the conclusion would be different.
Regards,
Hassan H.
08/02/13