Author Topic: Graphing an Absolute Value Function (Read 4362 times) Tweet Share

nerd · « **Reply #15 on:** November 05, 2008, 10:13:45 pm »

Daniel...i just corrected that in the last post

Mao · « **Reply #16 on:** November 05, 2008, 11:41:27 pm »

ooops, I'm an idiot

dusty_girl1144 · « **Reply #17 on:** November 05, 2008, 11:50:41 pm »

Quote from: nerd on November 05, 2008, 10:12:29 pm

Quote from: Mao on November 05, 2008, 08:19:30 pm
$|x-2|= \begin{cases} 2-x & x<2 \\ x-2 & x\ge 2 \\ \end{cases}$

$|x-4| = \begin{cases} 4-x & x<4 \\ x-4 & x\ge 4 \\ \end{cases}$

$\implies h(x)=|x-2|-|x-4|+1 = \begin{cases} -1 & x<2 \\ 2x-5 & 2\le x < 4 \\ 3 & x\ge 4 \\ \end{cases}$

Actually, Mao, i think you missread the question.

It asks for $h(x)=|x-2|-|x+4|+1$

$h(x)=|x-2|-|x+4|+1 = \begin{cases} -5 & x\ge 2 \\ -2x-1 & -4\le x \le 2 \\ 7 & x\le -4 \\ \end{cases}$

You have to look at the intercepts of the two absolute value functions and see where they intersect.

can someone run by me, how you got these?

Lulu · « **Reply #18 on:** November 06, 2008, 06:28:23 pm »

How would you do this one?

Define y = |x+2| + | x-2 | over the relevant intervals. Hence sketch the graph of y.

i'm just confused as to how you set up those hybrid functions and get the different curves for each section.

thanks

Mao · « **Reply #19 on:** November 06, 2008, 06:40:58 pm »

/0 · « **Reply #20 on:** November 06, 2008, 06:43:30 pm »

Ya $x=-2$ and $x = 2$ are your 'critical' points (for lack of a better word)

You gotta consider the intervals between them and outside of them and see how they affect tha absolute value of each part of your function

nerd · « **Reply #21 on:** November 06, 2008, 06:44:00 pm »

How do you know where to put the "less than or equal to" signs and where to put just the "less than" signs. Or doesn't it matter?

/0 · « **Reply #22 on:** November 06, 2008, 06:46:57 pm »

Quote from: nerd on November 06, 2008, 06:44:00 pm

How do you know where to put the "less than or equal to" signs and where to put just the "less than" signs. Or doesn't it matter?

You have to use the definition of the absolute value function.

$|x| = \begin{cases} x & x \geq 0 \\ -x & x< 0 \\ \end{cases}$

$|f(x)| = \begin{cases} f(x) & f(x) \geq 0 \\ -f(x) & f(x) < 0 \\ \end{cases}$

$f(|x|) = \begin{cases} f(x) & x \geq 0 \\ f(-x) & x < 0 \\ \end{cases}$

It's $\geq$ and $<$ .

Lulu · « **Reply #23 on:** November 06, 2008, 06:48:22 pm »

ok thanks for that mao, but with this step:

Quote from: Mao on November 06, 2008, 06:40:58 pm

$\begin{align*} |x+2|+|x-2| &= \begin{cases} -(x+2)-(x-2) & x<-2 \\ (x+2)-(x-2) & -2\le x < 2 \\ (x+2)+(x-2) & x \ge 2 \\ \end{cases} \\ y &= \begin{cases} -2x & x<-2 \\ 4 & -2\le x < 2 \\ 2x & x \ge 2 \\ \end{cases} \\ \end{align*}$

are you trying to find the intersection of the individual domains?

Captain · « **Reply #24 on:** November 06, 2008, 06:52:04 pm »

Quote from: Mao on November 05, 2008, 11:41:27 pm

ooops, I'm an idiot

Saved.

Thankyou

Mao · « **Reply #25 on:** November 06, 2008, 07:00:47 pm »

Lulu: yep

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Author Topic: Graphing an Absolute Value Function (Read 4362 times) Tweet Share

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