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Starting with Spivak - question

  1. Feb 28, 2015 #1
    Hey, I'm just starting out with calculus and am giving the Spivak book a try per threads on here. In the introduction to functions chapter spivak defines a function f(x) =x^2. for all x such that -17 ≤ x ≤ π/3
    Then says you should be able to check the following assertion about the function defined above:

    f(x+1) = f(x) + 2x + 1 if -17 ≤ x ≤ (π/3) - 1

    I don't understand why there a negative one is now needed at the end of the domain
  2. jcsd
  3. Feb 28, 2015 #2


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    Homework Helper

    -18 ≤ x ≤ (π/3) - 1 is the domain of f(x+1)
    -17 ≤ x ≤ (π/3) is the domain of f(x)
    -17 ≤ x ≤ (π/3) - 1 is the domain the two share

    edit:to correct error
    Last edited: Mar 1, 2015
  4. Feb 28, 2015 #3
    thanks for your reply,

    That doesn't make sense to me.

    My incorrect thinking goes like this..

    What is f(x+1) actually saying? To me its saying: apply the function f (which is to square) to the domain x (defined above) +1..making the domain of f(x+1) -16 to π/3 + 1
  5. Mar 1, 2015 #4
    Shouldn't the domain of f(x+1) be -18 ≤ x ≤ (π/3) - 1??
  6. Mar 1, 2015 #5


    Staff: Mentor

    The graph y = f(x + 1) is the graph of y = f(x) shifted (translated) one unit to the left. Since the domain of y = f(x) is -17 ≤ x ≤ π/3, the domain of y = f(x + 1) will be -18 ≤ x ≤ π/3 - 1. IOW, the original interval shifted one unit to the left.

    Forget the weird domain for a moment. You know what the graph of y = f(x) = x2 looks like, right? The graph of y = g(x) = f(x + 1) is a parabola the opens up, and whose vertex is at (-1, 0). Every point on the shifted parabola is one unit to the left of its corresponding point on the graph of y = f(x).

    Yes, which is different from what you said above.
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