Author Topic: Algebraically finding the domain (Read 4635 times) Tweet Share

kenhung123 · « **on:** March 23, 2010, 07:24:17 pm »

How do you find the domain of $\sqrt{x^2+x-6}$ ?
I tried $\sqrt{x^2+x-6}\geq 0$
$x\geq 3$ or $x\geq-2$ but it doesn't really make sense

the.watchman · « **Reply #1 on:** March 23, 2010, 07:27:21 pm »

Careful:

Let $x^2 + x - 6 \geq 0$

$(x+3)(x-2) \geq 0$

$x \geq 2$ OR $x \leq -3$

Blakhitman · « **Reply #2 on:** March 23, 2010, 07:28:50 pm »

I got $x\geq2$ and $x\geq-3$

Blakhitman · « **Reply #3 on:** March 23, 2010, 07:29:40 pm »

Lol he beat me....AND I made a mistake

.

the.watchman · « **Reply #4 on:** March 23, 2010, 07:30:25 pm »

Quote from: Blakhitman on March 23, 2010, 07:28:50 pm

I got $x\geq2$ and $x\geq-3$

If you're doing algebraic method, BE CAREFUL
You need to allow two cases:

CASE 1: (x+3) and (x-2) are BOTH GREATER THAN ZERO (or equal to) (to give a positive result)

CASE 2: (x+3) and (x-2) are BOTH LESS THAN ZERO (or equal to) (to give a positive result)

Safe way is to go graphically

m@tty · « **Reply #5 on:** March 23, 2010, 07:30:39 pm »

You require $x^2+x-6\geq 0$

$(x+3)(x-2)\geq0$

2 cases:
Both positive
$x+3\geq0 \cap x-2\geq0$

$x\geq2$

Both negative
$x+3\leq0 \cap x-2\leq0$

$x\leq-3$

$\therefore x^2+x-6\geq0 \ ,x\in\mathbb{R}\setminus(-3,2)$

Which is the same as the watchman's answer.

the.watchman · « **Reply #6 on:** March 23, 2010, 07:31:43 pm »

Quote from: m@tty on March 23, 2010, 07:30:39 pm

Which is the same as the watchman's answer.

YAY!

m@tty · « **Reply #7 on:** March 23, 2010, 07:32:25 pm »

You got it right for once

the.watchman · « **Reply #8 on:** March 23, 2010, 07:32:54 pm »

Quote from: m@tty on March 23, 2010, 07:32:25 pm

You got it right for once

Thanks a lot... >.<

Blakhitman · « **Reply #9 on:** March 23, 2010, 07:34:04 pm »

Quote from: m@tty on March 23, 2010, 07:32:25 pm

You got it right for once

Yeah I know, about time he is actually useful eh?

kenhung123 · « **Reply #10 on:** March 23, 2010, 07:39:20 pm »

Erm, so when doing these things, when the number is taken to the other side and becomes negative, then the arrow points the other way? When it is positive, it just stays the same?

Blakhitman · « **Reply #11 on:** March 23, 2010, 07:53:42 pm »

no you only change direction of arrow when you're multiplying by negative, so for example,

$-x\geq 26$

then it becomes

$x\leq -26$ by multiplying both sides by $-1$

What they mean is that there's two cases, as matty said

Quote from: m@tty on March 23, 2010, 07:30:39 pm

You require $x^2+x-6\geq 0$

$(x+3)(x-2)\geq0$

2 cases:
Both positive
$x+3\geq0 \cap x-2\geq0$

$x\geq2$

Both negative
$x+3\leq0 \cap x-2\leq0$

$x\leq-3$

$\therefore x^2+x-6\geq0 \ ,x\in\mathbb{R}\setminus(-2,3)$

Which is the same as the watchman's answer.

the.watchman · « **Reply #12 on:** March 23, 2010, 08:00:27 pm »

Quote from: Blakhitman on March 23, 2010, 07:53:42 pm

no you only change direction of arrow when you're multiplying by negative, so for example,

$-x\geq 26$

then it becomes

$x\leq -26$ by multiplying both sides by $-1$

Another way of saying this is an example:

$-x \geq 26$

Add x to both sides: $0 \geq x+26$

Subtract 26 from both sides: $-26 \geq x$ <=> $x \leq -26$

Sorry for the shitty explanation

EDIT: 1500 POSTS

Blakhitman · « **Reply #13 on:** March 23, 2010, 08:04:13 pm »

lol nah that's how I do it aswell, but I had to use the general thingy, you know, what people are taught haha.

oh and Kenhung watch this, very intuitive! http://www.youtube.com/watch?v=ZNtzWpU80-0&feature=related

TrueTears · « **Reply #14 on:** March 23, 2010, 08:06:00 pm »

Quote from: Blakhitman on March 23, 2010, 07:28:50 pm

I got $x\geq2$ and $x\geq-3$

if u did it this way the answer would simply be $x \ge 2$

but yeh im probs wrong

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