Graph q (1 Viewer)

[yashbb]

How do I know what the restriction is
Q9a

 

[ExtremelyBoredUser]

How do I know what the restriction is
Q9a

I don't think there's any restrictions assuming you're talking about discontinuities.

 

[yashbb]

I don't think there's any restrictions assuming you're talking about discontinuities.

Well it’s an absolute value graph and the answers show a restriction.

 

[yashbb]

To solve it you convert it to a piecewise function where one branch is x>=2 and the other would be x<2

So if x>=2 then the absolute value function becomes positive so |x-2| + x + 1 turns into (x-2) + x + 1 = 2x -1

If x<2 then the absolute value function becomes negative then |x-2| + x + 1 turns into -(x-2) + x + 1 = -x + 2 + x + 1 = 3

But I'm not sure what you mean by restriction?

Our teacher called it a restriction but what you have described is what I mean but I think one of your branches is wrong. One of them is x>=2

 

[ExtremelyBoredUser]

Well it’s an absolute value graph and the answers show a restriction.

No discontinuity?

Our teacher called it a restriction but what you have described is what I mean but I think one of your branches is wrong. One of them is x>=2

Oh I assume you're talking about domain?

 

[5uckerberg]

The question specifically asked for branches so in this case, they wanted us to say that when x is smaller than 2 what we will have is for and for . If you are talking about range then something is clear it will be .

 

[yashbb]

But it matches with the answer you provided...?

Sorry I can’t read lol

 

[yashbb]

The question specifically asked for branches so in this case, they wanted us to say that when x is smaller than 2 what we will have is for and for . If you are talking about range then something is clear it will be .

My q is why is that why is x smaller than 2 and larger than or equal 2. How do i work that out

 

[5uckerberg]

Not sure if it is a typo but do you mean x<2?

I do. mean , whoops, but yeah for it becomes as

 

[ExtremelyBoredUser]

s x smaller than 2 and larger than or equal 2. How do

It's a piecewise function so it can take on simultaneous properties such as x being less than 2 in a certain range.

The absolute value function can be decomposed into a piecewise function;

|x - 2| can be interpreted as and likewise from the domain as you can consider two to be the "turning point" (since its the x-intercept) [and for technicality, range is .

The regular graph:


Decomposed to piecewise



So that's how an abs. value function works. The question adds a regular equation "x+1". I assume that cambridge intends you to sketch these graphs and add the oordinates.



You would consider the x-intercept for the abs. value graph and the oordinate for x+1. You would get the new value when adding the graphs to be y=3. Then you pretty much add both sides, you can sort of imagine how the graph will end up like. x+1 is negative for values less than -1 and the abs. value function is increasing (y-axis) at the same rate. This is where the piecewise function helps;



So when adding the x+1 graph,


for

Likewise, you can picture the graph is only increasing to the right of the abs. value intercept as both x+1 and | x - 2 | is increasing at the same rate.

for (from the piecewise)


for

Hence you would get the new graph;



As you can see its pretty much the combination of both functions we got above.

for
for

Sorry I kinda realised I went too overdepth for a simple thing but just understanding abs. value graphs as piecewise functions made graphing way more easier, this is sorta my thought process.

 

[yashbb]

It's a piecewise function so it can take on simultaneous properties such as x being less than 2 in a certain range.

The absolute value function can be decomposed into a piecewise function;

can be interpreted asand likewise from the domain as you can consider two to be the "turning point" (since its the x-intercept) [and for technicality, range is .

The regular graph:
View attachment 33046

Decomposed to piecewise

View attachment 33048

So that's how an abs. value function works. The question adds a regular equation "x+1". I assume that cambridge intends you to sketch these graphs and add the oordinates.

View attachment 33050

You would consider the x-intercept for the abs. value graph and the oordinate for x+1. You would get the new value when adding the graphs to be y=3. Then you pretty much add both sides, you can sort of imagine how the graph will end up like. x+1 is negative for values less than -1 and the abs. value function is increasing (y-axis) at the same rate. This is where the piecewise function helps;



So when adding the x+1 graph,




Likewise, you can picture the graph is only increasing to the right of the abs. value intercept as both x+1 and | x - 2 | is increasing at the same rate.

(from the piecewise)




Hence you would get the new graph;

View attachment 33051

As you can see its pretty much the combination of both functions we got above.




Sorry I kinda realised I went too overdepth for a simple thing but just understanding abs. value graphs as piecewise functions made graphing way more easier, this is sorta my thought process.

Omg thx so much man, you once again have saved the day.