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Transforming functions

  1. Jun 16, 2015 #1
    1. The problem statement, all variables and given/known data
    If f(x)=|x-1/2|-5 determine g(x)=2f(-x+(3/2))

    2. Relevant equations

    3. The attempt at a solution
    Well, I tried to factor out the k-value in the g(x) formula.
    So I was left with:

    g(x)=2f(-1)(x-3/2)

    Then I multiply f(x) by 2 and am left with:
    g(x)=2|x-(1/2)|-10

    Then I subtract the 3/2 from 1/2 and am left with -2:
    g(x)=2|x-2|-10

    Then I apply the negative k-value and am left with
    g(x)=2|-x+2|-10


    I checked on desmos, and that answer is wrong. It should be:
    g(x)=2|-x+1|-10



    I've asked friends, looked online, in my textbook, and in my notes for things relating to this, and after 3 hours, came up empty-handed. If anyone could tell me where I went wrong. I would be very grateful. If you could also go step-by-step solving this problem, I would appreciate it.

    Thanks.
     
    Last edited: Jun 16, 2015
  2. jcsd
  3. Jun 16, 2015 #2

    Ray Vickson

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    To get g(x), replace x everywhere (in the f(x) formula) by -x + (3/2); after that, multiply the whole thing by 2. In other words, ##g(x) = 2 \left. f(t) \right|_{t = -x + 3/2}##.
     
  4. Jun 16, 2015 #3
    I don't follow the last bit. I think the absolute value sign got messed up. Thanks for responding!
     
  5. Jun 16, 2015 #4
    I get it now! Thanks so much! Now I can finally move on :smile:
     
  6. Jun 17, 2015 #5

    Ray Vickson

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    Just to be clear: the notation ##f(t)|_{t = u}## does NOT mean there is a missing absolute-value sign. The notation is shorthand for "##f(t)##, evaluated at ##t = u##". Of course, that is just ##f(u)##, but since you seemed to be confused by notation such as ##f(-x + 3/2)## (that is, where ##u = -x + 3/2##) I used the alternate notation instead. It is similar to the notation used in expressing definite integrals, such as
    [tex] \int f(x) \, dx = F(x) \Rightarrow \int_a^b f(x) \, dx = F(x)|_{x=a}^{b} = F(b) - F(a). [/tex]
     
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