1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    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!

Range of P(x)

  1. Oct 2, 2011 #1
    1. The problem statement, all variables and given/known data
    P(x) = √6+5x-x2


    2. Relevant equations



    3. The attempt at a solution
    P(x) = √6+5x-x2

    -x2+5x+6 >/= 0
    x2-5x-6 </= 0
    (x-6)(x+1) </= 0
     
  2. jcsd
  3. Oct 2, 2011 #2

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Do you mean [itex]P(x) =\sqrt{6+5x-x^2}\,?[/itex]

    If so, then use parentheses to make things clear.

    P(x) = √(6+5x-x2).

    Otherwise it looks like you have [itex]P(x) =(\sqrt{6})+5x-x^2\,?[/itex]

    What you have done will allow you to find the domain of P(x).

    To find the range, find the vertex of the parabola of f(x) = 6+5x-x2 . Then proceed from there.
     
  4. Oct 2, 2011 #3
    In vertex form it is y=(x-2.5)2-12.25
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Range of P(x)
  1. Substitute p=x^2 (Replies: 3)

  2. Finding range of f(x) (Replies: 13)

Loading...