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? Domain & Range ?

  1. Feb 16, 2008 #1
    Calculus Help, Please. (Domain, Range, Graphs)

    1. The problem statement, all variables and given/known data

    Determine its domain and range.
    f(x) = {2x+1, x<0
    {2x+2, x => 0

    f(x) = {x^2 + 2, x<=1
    {2x^2 + 2, x >1

    f(x) = { |x| + 1, x<1
    { -x +1, x=>1

    2. Relevant equations

    No equations necessary?

    3. The attempt at a solution

    I'm thinking I could graph it or solve the f(x) by plugging in the x value indicated from the question. I don't really know.

    1. The problem statement, all variables and given/known data

    Please see the attachment file at the bottom of this post called "Graph".
    The question is "Write the function whose graph is given in the figure."

    2. Relevant equations

    Slope equation:
    ( Y2 - Y1 ) / (X2 - X1 )

    3. The attempt at a solution

    (0 -(-1)) / (-2 -(-1)) = 1
    (-1 -1) / (3-(-1)) = -1/2

    So now I know the slopes of that might help me with getting the function. However, I'm stuck. Anybody know what to do next?

    Attached Files:

    Last edited: Feb 16, 2008
  2. jcsd
  3. Feb 16, 2008 #2


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    Homework Helper

    you don't need to post twice, it is forbidden.

    domain is what values of x you can have, and range is what values your function f can "reach".

    So start with the first one:
    f(x) = {2x+1, x<0
    {2x+2, x => 0

    What x's can we have? And which f(x) can we get?
  4. Feb 16, 2008 #3
    I posted twice due to the fact that I made a mistake on putting this thread here instead of putting it in the "Precalculus" Section.

    So, what you're saying is that if I plugged in:
    f(x) = 2x + 1, x<0
    = 2(-1) + 1 = 1
    = 2(-2) + 1 = 3
    = 2(-3) + 1 = 5
    = 2(-4) + 1 = 7

    f(x) = 2x+2, x=> 0
    = 2(0) + 2 = 2
    = 2(1) + 2 = 4
    = 2(2) + 2 = 6
    = 2(3) + 2 = 8

    So the answer should be domain is X>0 and the range is Y>0 ?
    I don't get it.
  5. Feb 16, 2008 #4
    Also, the answer from the book is the domain: (-infinity, infinity) and
    the range is (-infinity, 1), [2, infinity)

    I still don't get it.
  6. Feb 16, 2008 #5
    f(x)=2x+1, x<0.

    This ststement means that the fuction f(x) is defined (or exists) for all values of x which ate less than zero.

    Similarly f(x)=2x+2, x=>0 means that this function is defined for all values of x above and including 0.

    If you were to draw these you would have the first function all to the left of x=0 and the second function all above and at x=0.

    Since the graphs are straight lines try to think of the shape of the graphs as they go to their limits (most extreme x values) so for the first function negative infinity and zero. Whatever the y values approach, this will give you the range of tge function.

    This basically gives you your domain..
  7. Feb 16, 2008 #6
    (sorry meant negative infinty and 1)

    EDIT: no I didn't lol
    Last edited: Feb 16, 2008
  8. Feb 16, 2008 #7
    But, that doesn't make sense because 2(-1) + 1 = -1
    So, shouldn't it be negative infinity and -1 for the first function? Also, I graphed it as you said and this is what I came up with: (Please see the attachment called "Graph2" below this reply).

    I'm still confused on how the range is 1 and why the answer in the book is [2, infinity), what's that suppose to mean? Also, just for reference, I'm using "Calculus of A Single Variable -Early Transcendental Functions Third Editions" book.

    Attached Files:

  9. Feb 16, 2008 #8
    Why are you solving f(-1)? You need to solve for f(0) for the first function which gives 1. The other value should be negative infinity, I think you know this. So far the range is (-infinity,1) but this is only half of the question. You still have the second function. Remember this question is about a hybrid function (a function made of other functions).
  10. Feb 16, 2008 #9
    Now, I feel so dumb but isn't "X < 0" means that the "X" value need to be less than zero? So, less than 0 is -1 therefore the f(X) must be f(-1), right? I don't get why you have to solve for f(0) when it doesn't say X =< 0 to the first function. Or do I always have to assume to put "zero" first in the function and solve it that way THEN get the X value according to the rules indicated (eg. X < 0 )?

    Sorry for taking too much of your time. I just really want to get Calculus.
  11. Feb 16, 2008 #10
    Hold on. I think I got it.
    However, I still don't get my #2 Question:
    1. The problem statement, all variables and given/known data

    Please see the attachment file at the bottom of this post called "Graph".
    The question is "Write the function whose graph is given in the figure."

    2. Relevant equations

    Slope equation:
    ( Y2 - Y1 ) / (X2 - X1 )

    3. The attempt at a solution

    (0 -(-1)) / (-2 -(-1)) = 1
    (-1 -1) / (3-(-1)) = -1/2

    So now I know the slopes of that might help me with getting the function. However, I'm stuck. Anybody know what to do next?

    Attached Files:

  12. Feb 16, 2008 #11
    no you are right, you need to put a value less than zero. But their are smaller negative numbers which are closer to 0:-0.5,-0.1,-0.000000001. So where do we draw the line? Well we use a number which is infitely small and subtract it from 0, which basically gives zero. We say as x "approaches" zero.

    The way this is shown in your answer is by use of different brackets. Note for the first part the range is (-infinity,1) I.e. Using these "(,)" brackets. This means that the value you obtain is infinitely close to the value you put there, or another way of thinking of it is that it will only just not reach this value or will "approach" that value.

    Note also inthe second part of the function that the range is given [2,infinity) the square brackets show that the the function actually DOES reach this value.

    Also no need to be sorry haha.
  13. Feb 16, 2008 #12
    WOW! Thank you so much. Now I know why "[" is used. Plus, I get it! *clap*clap! You're much better than my Calculus teacher. =)
  14. Feb 16, 2008 #13
    remember though that with non-linear graphs the range might not be found by the extreme x-values. There may be a minimum/maximum value between the given domain. This is why if you cannot properly visualize the function it is always best to graph it. Even if you think you can visualise it it may still be beneficial to graph the function to be sure. Having it in front of you ofyen makes things easier.
  15. Feb 16, 2008 #14
    Oh, I'm also just wondering what does the "U" stands for? (Eg. (-infinity, 0) U (1, infinity)

    Edit: Thanks for the tip above. I'll keep that in mind.
  16. Feb 16, 2008 #15
    the "U" symbol represents the expression "in union with" but I suppose you can just say another very technical term.... "and"

    in this context it basically means the set of values from negative infinity to 0 "and" from 1 to infinity.
  17. Feb 16, 2008 #16
    Now for your second question!

    I think you have already realised this is a hybrid function with two parts. You have also correctly found the gradient (slope) of both of these parts. Good stuff!

    Now, as you can see the parts are linear functions and you know (i assume) that the general form of a linear function is of the type y=mx+c, where m is the gradient and c is the y-intercept.

    What you need to do is sub in the information you are given and tgen limit the domain for each part so it fits the function we want...

    So, for the first part lets use the point (-2,0) and the gradient you found to be 1.
    Given y=mx+c it follows that:
    therefore c=2.

    Now we arrive at y=x+2. However this is still not correct because y=x+2 has a domain of all real numbers whereas in the graph you provided tgis part of the hybrid function clearly does not. We are going to have to "restrict" the domain.

    You can see from your graph the values which y=x+2 does exist, this is from (-2,-1) so we say that this part of the hybrid function follows the equation:

    y=x+2, -2<x<-1

    Note that I have used < not =< because the points you gave were in () brackets, if they were in [] brackets you would need to use <=.

    Now try the second part and when youbare done write the final hybrid function as those in your first question.
  18. Feb 16, 2008 #17
    Thanks for the reply. Now, I follow all except how you have gotten the y-value of -1. Isn't suppose to be 0 since that -2 + 2 = 0? Also, the book's answer is -2<=x<=-1. Do you think they might have made a mistake? Because my teacher said that there are few mistakes in the book.

    I'll try the second part now from what you've told me:

    (3, -1)
    -1 = -1/2(3) + c
    c = 1/2

    and then I'm stuck again because I have no idea how to get the restriction.
    Last edited: Feb 16, 2008
  19. Feb 16, 2008 #18
    Hold on, now that I think about it, do I just have to look at the graph to determine the restriction? Since the graph (referring to the first function) does not extend to the x-value (domain) of -2 and -1, does that mean that that's the answer? It's between those points? But, I still don't get it since the book says -2<=x<=-1. How can I determine when to put the "equal" thing in the restriction?

    Also, it's kind of the same thing with the second function since the graph of x-values does not extend to 3 and -1 does that mean that -1<x<3? But, the book says that the answer is -1<x<=3. I simply don't get the fact when to put the "equal" sign in the restriction area. Please help.

    I feel so dumb. =(
  20. Feb 16, 2008 #19
    im not sure what you mean about -2+2=0... How are you getting this?

    If the books answer uses "<=" etc. then the graph either uses [] brackets for the endpoints or the lines have filled in black dots at the endpoints. If there are filled in black dots this means the same as square brackets, that the equation DOES exist for the said values. The alternative is hollow dots "o" at the end of the line segments, this is equivalent to () brackets, meaning that the equation APPROACHES the given value.

    Perhaps your book has this but you did not add the detail to your graph?

    Your equation y=-1/2x+1/2 is correct. Now domain is what x-valures the equation is defined (exists) over. So if you look at the graph. What are it's most extreme x-values?
  21. Feb 16, 2008 #20
    For the -2+2=0.. You know how the equation is y=x+2, I just assumed that since we have the x value from the first function of -2, we could just plug it in there to find the y-value. Therefore making it (-2, 0). But, obviously, it's wrong because the x values are in between the -2 and -1.

    Yes, you're right. I definitely did not put the filled black dots on the graph. All of the points should have had filled black dots on there. However, if that is the case, I'm still confused due to the fact that the answer to the second function is -1<x<=3. How come there's a < instead of it being <= like everything else? Because as you have said, if the filled black dots are there, everything should have that equal sign in the restriction.

    So what went wrong?
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