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Simple graphing

  1. Jan 10, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]f(x)[/tex]= -x-1,-4[tex]\leq[/tex] [tex]x[/tex] [tex]\leq[/tex]-1
    ......= x+1, -1[tex]\leq[/tex] [tex]x[/tex] [tex]\leq[/tex]0
    ......= -x+1, 0[tex]\leq[/tex] [tex]x[/tex] [tex]\leq[/tex]4

    obtain the graph of [tex]g(x)=3f(1/2(x+2))-2[/tex] from the graph of [tex]f(x)[/tex]


    3. The attempt at a solution

    i am not sure how to obtain the graph.

    for example, should i plug in (1/2(x+2)) into -x-1, and have something like, -3(1/2(x+2))-2-1.
     
    Last edited: Jan 10, 2008
  2. jcsd
  3. Jan 10, 2008 #2

    HallsofIvy

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    First, of all you can't graph f itself- it is not a function! In particular, from the first line, f(-1)= -1-1= -2 while, from the second line, f(-1)= -1+ 1= 0. One of those should be "[itex]<[/itex]" rather than "[itex]\le[/itex]". I will assume one of each of those is "<" but you will have to determine exactly which point goes in which interval.

    Also, do you mean (1/2)(x+2) or 1/(2(x+2))? I will assume the former.

    f(x) is "piecewise linear" and the argument of f, (1/2)(x+2) is linear so you can just determine for what values of x, in (1/2)(x+2) give a value in each of the intervals used in defining f(x).

    If (1/2)(x+ 2)= -4, then x+ 2= -8 so x= -10. If (1/2)(x+ 2)= -1, then x+ 2= -2 so x= -4. Since everything is linear here, g(-10)= 3f(-4)-2= 3(-5)- 2= -17 while g(-4)= 3f(-1)- 2= 3(-2)- 2= -8. The first part of the graph of g(x) is a straight line from (-10, -17) to (-4, -8).

    As before, x= -4 gives (1/2)(x+ 2)= -1 and, here, f(-1)= 0. so g(-4)= 3(0)- 2= -2. If (1/2)(x+ 2)= 0, then x+ 2= 0 so x= -2. g(-2)= 3f(0)- 2= 1. The next part of the graph of g is the straight line from (-4, -2) to (-2, 1).

    As before, x= -2 gives (1/2)(x+ 2)= 0 and, here, g(-2)= 3f(0)- 2= 3(1)- 2= 1 also. (1/2)(x+ 2)= 4 gives x+ 2= 8 so x= 6. g(6)= 3f(4)- 2= 3(-3)-2= -11. The final part of the graph of g is the straight line from (-2, 1) to (6, -11).
     
  4. Jan 11, 2008 #3
    its -x-1, so -(-1)-1=0 and x+1, -1+1=0

    it is a function.

    what i dont understand is how i am to obtain the graph from f(x). i get the correct f(x). i just don't know if i am to transform it or do some other bs thing to it.

    and yes it is (1/2)(x+2).

    what does 3f((1/2)(x+2)) mean??? i dont understand this.

    your explanation is confusing me somewhat."the argument of f, (1/2)(x+2) is linear so you can just determine for what values of x, in (1/2)(x+2) give a value in each of the intervals used in defining f(x)." i dont understand this.
     
  5. Jan 11, 2008 #4
    If you have a function [tex] f(y) = y + 1 [/tex] then the meaning of [tex] f \left( \frac{1}{2} (x+2) \right) [/tex] is [tex] f(y) [/tex] when [tex] y = \frac{1}{2} (x+2) [/tex] and thus

    [tex] f( y = \frac{1}{2} (x+2) ) = \frac{1}{2} (x+2) + 1 [/tex]
     
    Last edited: Jan 11, 2008
  6. Jan 11, 2008 #5

    HallsofIvy

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    Dang, it's my eyes again!

    What part do you not understand? There are quite a number of things said there. Do you understand that (1/2)(x+2) is linear? The graph of a linear function, between x= a and x= b is just the straight line between (a,f(a)) and (b, f(b)). You can determine that just by calculating the two end points.
    If f(y) is given between a and b, and y= some function of x, then you have to figure out what values of x correspond to y= a and y= b.
     
  7. Jan 12, 2008 #6
    i understand that the (1/2)(x+2) is linear as x is to the power of 1.

    the bolded part is what is confusing me.
     
  8. Jan 12, 2008 #7
    as tried to understand your explanation. i think i get some clearness.

    [tex]g(x)=3f(1/2(x+2))-2[/tex] i think this means that put a value in for x. say x= -4. then that makes it f(-1). so then i look at my orignal graph. and see what value for f(-1) is there.
    so since its 0. if i multiply it by 3 nothing happens and then i just move it down by 2 and it becomes -2.

    by the y=a and y=b, i think that means from -4 to 4 as the graph of f(x) exists only in that interval and so must g(x).

    am i correct in this extraction??
     
  9. Jan 12, 2008 #8

    dynamicsolo

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    You can carry out the transformations of f(x) into g(x) algebraically with some effort (although it's giving me a bit of a headache just to read this problem).

    I suspect what the person posing this problem may want you to do is make the transformations "graphically". Have you had all that stuff about what happens to the graph of a function f(x) when it becomes f(x+2), when it becomes f( [1/2]·(x+2) ) , etc.? (All that horizontal and vertical "shifts" and "stretches" and "squashes" jazz?) It may be all that they're looking for is to have you draw the picture of f(x), perform the graphical transformations in the appropriate sequence, and obtain g(x) at the end of that.

    If you've had the stuff I'm talking about, it's much less work than calculating what happens to each branch of the definition analytically and then having to plot that. (In fact, I think the problem was deliberately couched this way to drive you to use the graphical technique...)
     
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