How am I to find a range of a function?

What if I had y = |5-2x| for the same domain? So, y = |5-2(0)|, y = |5-2(-3)|, y = |5-2(4)|. I do not get a absolute value looking graph
 
Kulla_9289 said:
I don't know how to write this as a piecewise function. I know how to write y = |5-2x| as a piecewise function.
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Oh for goodness sake.

[math]f(x) = 5 - |2x| \text { if } - 3 \le x \le 4 \implies \\ f(x) = 5 - (-2x) = 5 + 2x \text { if } - 3 \le x < 0 \text { and }\\ f(x) = 5 - 2x \text { if } 0 \le x \le 4.[/math]
 
Kulla_9289 said:
@Dr.Peterson so my range: -3≤x≤5. But how would I know that the function changes behaviour at 0?
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The key idea for absolute values is that the absolute value changes behavior when its argument is zero, as I said:
Dr.Peterson said:
First, observe that the function changes its behavior when we take the absolute value of 0, namely at x=0.
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This is because of what you were taught about writing the absolute value function itself as a piecewise function.

Similarly, for |5-2x|, when does 5-2x equal zero?

This is how you write such functions piecewise, and also how you graph them. Go back and read the material you sent us, from the start; they teach you all about this.
 
JeffM said:
Oh for goodness sake.

[math]f(x) = 5 - |2x| \text { if } - 3 \le x \le 4 \implies \\ f(x) = 5 - (-2x) = 5 + 2x \text { if } - 3 \le x < 0 \text { and }\\ f(x) = 5 - 2x \text { if } 0 \le x \le 4.[/math]
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Why do the domain change?
 
Kulla_9289 said:
The domain inside the piecewise changed?
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The domain of the original function was [imath][-3, \ 4][/imath], right?

The domain of the equivalent function defined piecewise is [imath][-3, \ 0) \ \bigcup \ [0, \ 4] \equiv [-3, \ 4].[/imath]
 
How and why? I think you should start from scratch. I am only familiar with y = |5-2x|. So, f(x) = {5-2x 5-2x≥0 and -5+2x 5-2x<0}
 
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Kulla_9289 said:
@Dr.Peterson now we are talking. Why does a 5-2x = 0?
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Are you asking why I want to solve 5-2x = 0 in order to analyze |5-2x|, or how to solve it? Your English is imperfect, so it's hard to be sure what you mean.
Kulla_9289 said:
How and why? I think you should start from sketch. I am only familiar with y = |5-2x|. So, f(x) = {5-2x 5-2x≥0 and -5+2x 5-2x<0}
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That's right. So can you solve 5-2x≥0 and 5-2x<0? What do you get?

(The usual phrase is "start from scratch", which means "start from the beginning". Or do you mean, start by sketching the graph?)
 
Kulla_9289 said:
I meant why has 5-2x been equated to 0?
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Have you thought about it at all yourself? We're trying to work with you, so we need to know what you are thinking in order to communicate at the needed level.

If you wrote
Kulla_9289 said:
y = |5-2x|. So, f(x) = {5-2x 5-2x≥0 and -5+2x 5-2x<0}
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then you presumably know the answer to your own question. Why did you write 5-2x≥0? The reason for that is the same as for the equation, which determines the boundary of the two "pieces"; that is exactly what I mean by "changes behavior".

Then the next question for you will be, why did you not solve those two inequalities?
 
Kulla_9289 said:
-2x ≥ -5 so x ≤ 5/2 and x > 5/2. Overall, f(x) = {5-2x x ≤ 5/2 and -5+2x x > 5/2}
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Good.
Kulla_9289 said:
I equated it to zero because |x| is defined as: |x| = {x x≥0 and -x<0}.
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Good.

Now please tell us what you still don't understand, if anything.

And after you tell us what you don't understand, show us your attempt at solving whatever problem you are still working on.
 
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