Piecewise function

  • #1
321
20

Homework Statement


Hello I uploaded a picture of the image.

Find an expression for a function who's graph is the given curve.

Homework Equations




The Attempt at a Solution


The first function [0,3] y= -x+3 and the second one is y=2x-6 from 3 to 5

what I dont get, is why isnt the second one [3,5], instead the answer is (3,5)..

The way I see it at that point, x=3, it can be any one of those equations...
 

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Answers and Replies

  • #2
321
20
My thinking is because the function goes from right to left it would make sense that the first function is less than or equal to 3 but I'm not sure if thats right..
 
  • #3
34,152
5,768

Homework Statement


Hello I uploaded a picture of the image.

Find an expression for a function who's graph is the given curve.

Homework Equations




The Attempt at a Solution


The first function [0,3] y= -x+3 and the second one is y=2x-6 from 3 to 5

what I dont get, is why isnt the second one [3,5], instead the answer is (3,5)..

The way I see it at that point, x=3, it can be any one of those equations...
And that's why you should choose which function goes with which interval. If ##x \in [0. 3]## you use the formula y = -x + 3. If ##x \in (3, 5)##, you use the formula y = 2x - 6.
 
  • #4
321
20
And that's why you should choose which function goes with which interval. If ##x \in [0. 3]## you use the formula y = -x + 3. If ##x \in (3, 5)##, you use the formula y = 2x - 6.
So the reason my books answer was [0,3] for y=-x + 3 was because they chose it to be that.

But if I wanted to I could choose [0,3) for y=-x+3 and [3,5] for 2x - 6 and my answer would still be valid because that's what I decided to choose?
 
  • #5
34,152
5,768
So the reason my books answer was [0,3] for y=-x + 3 was because they chose it to be that.

But if I wanted to I could choose [0,3) for y=-x+3 and [3,5] for 2x - 6 and my answer would still be valid because that's what I decided to choose?
Yes. At x = 3, both functions have y-values of 0, so it doesn't matter how you define the endpoints of the two intervals.
 

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