# Thread: how do i solve question 1? and is the answer for qusetion 2 correct?

1. ## how do i solve question 1? and is the answer for qusetion 2 correct?

1)The domain of f(x)=

2)

a)x=2
b)x=4
c)x=-2
d)3

my answer is a for both qusetions is that right?

2. 1. $D_p : {(-\infty , -3)}\cup{(-3,4)}\cup{(4, +\infty)}$

2: $x=2$

3. Originally Posted by yeKciM
1. $D_p : {(-\infty , -3)}\cup{(-3,4)}\cup{(4, +\infty)}$

2: $x=2$
thankx for ur reply ....for question 1 what is the answer yekcim??and what is Dp??

4. that's region where ur function is defined.... u probably don't mark it as $D_p$ sorry
meaning that ur function is not defined in -3 and 4 ... and everywhere else it is...

5. Originally Posted by yeKciM
that's region where ur function is defined.... u probably don't mark it as $D_p$ sorry
meaning that ur function is not defined in -3 and 4 ... and everywhere else it is...
yeah but this is a multiple choice question where i have to choose only 1 answer... so whats the right answer??

6. sorry ... didn't look at this right... ur function is $f_{(x)}=\frac{x-4}{(x+3)(x-4)}=\frac{1}{x+3}$ so it's just -3 sorry sorry

it should be a) R{-3}

$\frac{x- 4}{x^2- x- 12}= \frac{x-4}{(x-4)(x+3)}$
is not defined for x= 4 or x= -3.

$\frac{x- 4}{(x- 4)(x+ 3)}= \frac{1}{x+3}$ only for $x\ne 4$.

If x= 4, $\frac{x- 4}{(x- 4)(x+ 3)}= \frac{0}{0}$ which is indeterminant while $\frac{1}{x+ 3}= \frac{1}{7}$.

8. yes i know but when u plot it ... it's define in x=4....
just for -3 it's not

9. Originally Posted by yeKciM
yes i know but when u plot it ... it's define in x=4....
just for -3 it's not
ok so i'm confused now?? !! what is the answer which choice is it a,b,c or d??

10. for the rational function $\displaystyle f(x) = \frac{x-4}{(x+3)(x-4)}$ , there is a vertical asymptote at $x = -3$ and a point discontinuity at $x = 4$.

11. Originally Posted by soso4
ok so i'm confused now?? !! what is the answer which choice is it a,b,c or d??
well i think its just -3

reason why i think that is, if we look at function like

$f_{(x)}=\frac{x-4}{(x-4)(x+3)}$

it would mean that it's not defined in $x=4 , x=-3$ and that it has stationary point in $x=4$ but that isn't true ... it hasn't any stationary point's and it's defined in $x=4$ as u see from plot that i posted .... witch is actually function

$f_{(x)}=\frac {1}{x+3}$

lol... didn't have this much of polemics for such a simple function a long time

P.S. u shouldn't be just interested in "which choice is it a,b,c or d??"... it's more important to u to understand how and why does i say this, or HelsofIvy that, or skeeter... rather then be just interested in right answer... and soon have more problems (with same type) which u can't solve

12. Originally Posted by yeKciM
well i think its just -3
and you are incorrect.

note the following graphs ...

the first graph is a large view of the function showing the vertical asymptote.

the second graph is the function "blown up" near x = 4 ... note the "hole", a point discontinuity.

13. Originally Posted by skeeter
and you are incorrect.

note the following graphs ...

the first graph is a large view of the function showing the vertical asymptote.

the second graph is the function "blown up" near x = 4 ... note the "hole", a point discontinuity.
in which app did u plot it ?
i zoom my a loot and it doesn't show that ..
lol I'm not saying that ur wrong just need to know which app do u use

14. TI-84 emulator

15. Originally Posted by yeKciM
yes i know but when u plot it ... it's define in x=4....
just for -3 it's not
No, it isn't. The graph of $y= \frac{x-4}{(x-4)(x+3)}$ looks like a hyperbola but with a hole at (4, 1/7). The graph of y= 1/(x+3) is the same hyperbola without the hole. If you plotted it with a calculator or a computer graphing program, your grid was probably too course, and it "jumped over" x= 4.

It can be crucially important in Calculus to distinguish things like $\frac{x-4}{(x-4)(x+3)}$ from $\frac{1}{x+3}$.

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