Plot some points and you with see what it looks like, it's actually really simple when you think about it.
For question 12 (v)
How do you find the domain and range for that graph? I don't even know how to draw that type of absolute value graph lol
Any help would be appreciated
Thanks
ohhh lol thx i never thought about plotting points now i get itPlot some points and you with see what it looks like, it's actually really simple when you think about it.
Thanks your posts was really helpful lol thx for redefining functions 4 meDomain: All x
How do I know this?
-> The Absolute Value function does not restrict what X values I can put in, it only restricts what Y values come out.
Think back to the very basics of functions, functions are not 'graphs' graphs are merely visual representations of functions.
A function is somethng where if you input 1 value you get ONE value.
Multiple inputs can have 1 output though. This might sound confusing but it is important to understand the concept.
So let us look at the Absolute Value function, does it restrict our input? No.
Is there a functin that does restrict input? Yes.
Look at for instance.
We are restricted to only non-negative inputs.
We know x=/=0 since we can't divive by zero, and we know that our input into f(x) is not restricted, hence we get
All real x, x=/=0
Range:
y=1, y=-1
This seems strange, have a range only applicable for two values, unlike what we normally have as infinite values. (i.e. -1 < y < 1)
But that's fine, f(x) can only ever be 1, or -1. Why?
Because of the definition of |x|
If you remember:
So lets put this definition into action.
So for x > 0
f(x) = x / x = 1
for x < 0
f(x)= x / -x = -1
Hence we can only ever get two outputs, therefore the range is y=-1, y=1
You can certainly draw a table of values to get a picture of what the graph looks like. However, it is not the most efficient way of doing it.domain: all real x, x not equal to 0
range: all real y, y not equal to 0 EDIT: I'm wrong lol, sy is correct for this
that's what I think it is, I could be wrong, basically you just sub in x values for domain that are possible and not like imaginary numbers
for range you just get the range of y values possible from the possible or real x values
I dunno how to draw this graph but I would plot a table of values lol and then draw out the graph? dunno if this is the right way
domain: all real x, x not equal to 0
range: all real y, y not equal to 0 EDIT: I'm wrong lol, sy is correct for this
that's what I think it is, I could be wrong, basically you just sub in x values for domain that are possible and not like imaginary numbers
for range you just get the range of y values possible from the possible or real x values
I dunno how to draw this graph but I would plot a table of values lol and then draw out the graph? dunno if this is the right way
Yea thanks guise appreciate itYou can certainly draw a table of values to get a picture of what the graph looks like. However, it is not the most efficient way of doing it.
The best way is to recall the 'formal' definition of the absolute value and take care of the conditions as Sy has pointed out. If you do this, you get y=1 for x>0 and y= -1 for x<0.
As show below:
Just find g(x) (which we can since we are given two points)Yea thanks guise appreciate it
Got another question lol:
I dont even know what question 4 is asking?
Thanks
Just find g(x) (which we can since we are given two points)
Then state range
ahh okai thanks guiseConsider some linear functions e.g. y=2x+5 , y=5x+1, y=6x.
Notice that all linear functions can be written in slope intercept form i.e. y=mx+b. However for these questions I have a strange tendency to say linear functions are in the form y=ax+b. So i write 'a' instead of 'm'....not that it matters.
So you need to develop a pair of equations using g(3)=2 and g(4)=4. So begin like this:
Linear equations are of the form g(x)=ax+b.
When x=3, g(3)=2 so:
2=3a+b --------(1)
When x=4, g(4)=4 so:
4=4a+b --------(2)
Solve these two equations simultaneously to find 'a' and 'b'. Sub them back into g(x)=ax+b and that is your answer.