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Homework Help: Range of a Function

  1. Mar 18, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the range of h

    2. Relevant equations

    h(x) = sqrt(25 + (x - 3)2)

    3. The attempt at a solution

    I factored out the (x-3)2 and simplified to get

    sqrt(x2 - 6x + 34)

    I was trying to figure out the domain first, knowing that x2 - 6x >= -34 in order for the number inside the sqrt to be non-negative.. I don't know, maybe I'm missing something really basic here. Help? :)
     
  2. jcsd
  3. Mar 18, 2010 #2

    Mark44

    Staff: Mentor

    Let's get the terminology straight. You expanded (x-3)2 ; it was already factored.
    The best thing, IMO, was to leave the radicand in its given form, 25 + (x-3)2. Looking at that as its own function, what is the range of this function? That will tell you a lot about the range of h(x).
     
  4. Mar 18, 2010 #3
    Well the range of 25 + (x - 3)2 is y >= 25, right? So, the range of sqrt(25 + (x - 3)2 is y >= 5?

    Yea, I think expanding the (x - 3)2 term messed me up. I was thinking that i was possible for the radicand to be a negative number.. ugh. Stupid mistake.

    Thanks a lot for the help.
     
  5. Mar 18, 2010 #4

    Char. Limit

    User Avatar
    Gold Member

    Also, just to let you know...

    x^2-6x has a minimum of -9. It will never be less than -34.
     
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