# Thread: Points not included in Domain

1. ## Points not included in Domain

image >>>

For 21

book gives answer: $x \in (2,4)$
but the function at x = 2 becomes $\sqrt{\frac{0}{-8}} \implies \sqrt{0}$
so shouldn't point 2 be included in the domain?

For 23

book gives answer: $x \in (2,infinity)$

but here also at x=2 function is defined

i think in both questions point 2 should be included ... just need confirmation or reason why it is not included?

2. $\displaystyle x^2-2x-8 = (x-4)(x+2) \implies x \neq -2,4$

3. Its 2 not -2

4. Maybe the book's answer is incorrect.

5. Hello, cupid!

$21.\;f(x) \;=\;\sqrt{\dfrac{\log_{0.3}(x-1)}{x^2-2x-8}}$

First, note that: . $x\,>\,1$

. . [1] The numerator and denominator are both positive, or
. . [2] The numerator and denominator are both negative.

Case [1]

$\log_{0.3}(x-1) \:>\:0 \quad\Rightarrow\quad x-1 \:>\:0.3^0 \quad\Rightarrow\quad x \:>\:2$

$(x+2)(x-4) \:>\:0 \quad\Rightarrow\quad x \,<\,\text{-}2\,\text{ or }\,x \,>\,4$

. . Hence: . $x\,>\,4$

Case [2]

$\log_{0.3}(x-1)\,\le\,0 \quad\Rightarrow\quad x-1\,\le\,0.3^0 \quad\Rightarrow\quad 1\,<\,x \,\le\,2$

$(x+2)(x-4)\:<\:0 \quad\Rightarrow\quad -2 \,<\,x\,<\,4$

. . Hence: . $1\,<\,x\,\le\,2$

$\text{The domain is: }\;\left(1,2\,\right] \cup (4,\infty)$

6. your domain include point 6

so $log_{0.3}5 = -1.337$
and denominator is 16

so root will be of negative number
which is not defined (we're not talking about complex numbers)

so this domain is wrong

right now i have to go to my class ... will post my solution later.

7. Originally Posted by cupid
so $log_{0.3}5 = -1.337$
and denominator is 16 so root will be of negative number
which is not defined (we're not talking about complex numbers)
so this domain is wrong.
The domain is indeed $[2,4)$.
I suspect whoever supplied the book's answer was just careless.
If you graph the function, it is easy to see this.

8. can you provide me with the the graph ??

And what about the 23rd ?

9. Originally Posted by cupid
can you provide me with the the graph ?
.

10. Originally Posted by Soroban
Hello, cupid!

First, note that: . $x\,>\,1$

. . [1] The numerator and denominator are both positive, or
. . [2] The numerator and denominator are both negative.

Case [1]

$\log_{0.3}(x-1) \:>\:0 \quad\Rightarrow\quad x-1 \:>\:0.3^0 \quad\Rightarrow\quad x \:>\:2$

$(x+2)(x-4) \:>\:0 \quad\Rightarrow\quad x \,<\,\text{-}2\,\text{ or }\,x \,>\,4$

. . Hence: . $x\,>\,4$

Case [2]

$\log_{0.3}(x-1)\,\le\,0 \quad\Rightarrow\quad x-1\,\le\,0.3^0 \quad\Rightarrow\quad 1\,<\,x \,\le\,2$

$(x+2)(x-4)\:<\:0 \quad\Rightarrow\quad -2 \,<\,x\,<\,4$

. . Hence: . $1\,<\,x\,\le\,2$

$\text{The domain is: }\;\left(1,2\,\right] \cup (4,\infty)$

Here's my solution >>> PIC
[img][/img]