1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:


    Then I multiply f(x) by 2 and am left with:

    Then I subtract the 3/2 from 1/2 and am left with -2:

    Then I apply the negative k-value and am left with

    I checked on desmos, and that answer is wrong. It should be:

    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.

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

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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]
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted