I seriously in this functions question Chop chop

[lilsheltie]

hey guys, I'm new here but whatever, who cares.
ANYWAYS, I have this math question on hybrid functions.
I desperately need help for this cause I'll be having an important test next week.
HELP!

OK anyways, in hybrid function,
example:

this question asked me to form the function by giving me a hybrid functions graph.
So i managed to find f(x) but I had a problem understanding the domains.

2x-1 , x<0
f(x)= 3x²-2x , 0≤x≤2
4 , x>2​

ok my question is.. for the domain, how do you know if you put ≤ instead of < ? HOw come it is not x≤0 and 0<x≤2 instead?

PS: well if you don't really know how to explain, or don't really understand what I am trying to ask, then, it'll be helpful if you could provide me a link to a website I can go to for help on hybrid functions.

Thanks! (=

 

[James R]

At x=0, the expression 2x-1 has value -1.
At x=0, the expression 3x^2 - 2x has value 0.

For the composite function, you need to decide which value you want at x=0. It seems from above that you have chosen the function to have the value 0 at x=0.

See?

By the way, on a graph, where there is a discontinuity it is usual to put a closed or open dot to indicate the value at the discontinuity. So, for the function above, the 2x-1 line would have an open dot at the point (0,-1), while the 3x^2 - 2x curve would have a closed dot or circle at the point (0,0).

 

[lilsheltie]

At x=0, the expression 2x-1 has value -1.
At x=0, the expression 3x^2 - 2x has value 0.

For the composite function, you need to decide which value you want at x=0. It seems from above that you have chosen the function to have the value 0 at x=0.

See?

By the way, on a graph, where there is a discontinuity it is usual to put a closed or open dot to indicate the value at the discontinuity. So, for the function above, the 2x-1 line would have an open dot at the point (0,-1), while the 3x^2 - 2x curve would have a closed dot or circle at the point (0,0).

ok if u say that, this question comes from my textbook

Sketch the graph of the function

-2x-2 ,x<0
f(x)= x-2 , 0≤x<2
3x-6 , x≥2

how come at x=0, both expressions -2x-2 and x-2 has the same value at -2?

 

[James R]

ok if u say that, this question comes from my textbook

Sketch the graph of the function

-2x-2 ,x<0
f(x)= x-2 , 0≤x<2
3x-6 , x≥2

how come at x=0, both expressions -2x-2 and x-2 has the same value at -2?

Since in that case the function is continuous at x=0, you could just as easily write:

-2x-2 ,x≤0
f(x)= x-2 , 0≤x<2
3x-6 , x≥2

or even

-2x-2 ,x≤0
f(x)= x-2 , 0<x<2
3x-6 , x≥2

It would make no difference in this case.

 

[dextercioby]

You can ask the same question for the point x=2...It means that the function is continuous in the 2 points...I don't know if you know what that means,but it's useful to know that it's not essential for this problem...So basically,u have to draw 3 lines and that's it.Can u do that...?

Daniel.

 

[lilsheltie]

At x=0, the expression 2x-1 has value -1.
At x=0, the expression 3x^2 - 2x has value 0.

For the composite function, you need to decide which value you want at x=0. It seems from above that you have chosen the function to have the value 0 at x=0.

See?

By the way, on a graph, where there is a discontinuity it is usual to put a closed or open dot to indicate the value at the discontinuity. So, for the function above, the 2x-1 line would have an open dot at the point (0,-1), while the 3x^2 - 2x curve would have a closed dot or circle at the point (0,0).

erm, how did u know that
For the composite function, you need to decide which value you want at x=0. It seems from above that you have chosen the function to have the value 0 at x=0.

 

[lilsheltie]

dextercioby: I seriously have zilch idea on what you're trying to say?

 

[dextercioby]

Perfect,at least can u draw tha graph...?

Daniel.

 

[lilsheltie]

I can but it's not the drawing of the graph that I am having a problem with, it's the... detestable signs. ARGHH!

 

[dextercioby]

What do you mean signs...?It's just + and -...There's nothing complicated about it.Basically u'll have to plot the function on each interval.

Daniel.

 

[James R]

lilsheltie:

Let me give you a simple example.

Suppose we have

f(x) = 3,\qquad x \geq 1
f(x) = 7, \qquad x &lt; 1

What is the value of f(x) at x=1? Well, which interval contains x=1? Obviously, the interval x &lt; 1 doesn't include x=1, so that part of the function definition doesn't apply. On the other hand, the other part of the function does include x=1. Therefore, for this function f(1) = 3, and f(1) definitely doesn't equal 7.

Now compare the function in your original question, at the point x=0. Which interval contains x=0?

 

[lilsheltie]

Since in that case the function is continuous at x=0, you could just as easily write:

-2x-2 ,x≤0
f(x)= x-2 , 0≤x<2
3x-6 , x≥2

or even

-2x-2 ,x≤0
f(x)= x-2 , 0<x<2
3x-6 , x≥2

It would make no difference in this case.

ok. I didn't know that.

 

[lilsheltie]

lilsheltie:

Let me give you a simple example.

Suppose we have

f(x) = 3,\qquad x \geq 1
f(x) = 7, \qquad x &lt; 1

What is the value of f(x) at x=1? Well, which interval contains x=1? Obviously, the interval x &lt; 1 doesn't include x=1, so that part of the function definition doesn't apply. On the other hand, the other part of the function does include x=1. Therefore, for this function f(1) = 3, and f(1) definitely doesn't equal 7.

Now compare the function in your original question, at the point x=0. Which interval contains x=0?

Huh? Ok I've given you quite a few examples so I am pretty much confused.
Anyways, I kinda get the picture now! (=