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Homework Help: Inequality problem

  1. Jun 25, 2010 #1
    1. The problem statement, all variables and given/known data

    Prove that: [tex]y^5+y^2-7y+5\geq 0[/tex] ,for all [tex]y\geq 1[/tex]



    2. Relevant equations



    3. The attempt at a solution

    [tex]y^5\geq 1[/tex] and [tex]y^2\geq 1[/tex] => [tex]y^5+y^2\geq 2[/tex].

    Also [tex]-7y+5\leq -2[/tex] , and then?
     
  2. jcsd
  3. Jun 25, 2010 #2

    Mark44

    Staff: Mentor

    If f(y) = y5 + y2 - 7y + 5, note that f(1) = 0.

    Look at f'(y) to determine where f is increasing and decreasing for y >= 1.
     
  4. Jun 25, 2010 #3
    O.K

    f'(y)=[tex]5y^4+2y-7\geq 0[/tex] for [tex]y\geq 1[/tex] .

    But how can this effect our f(y) ??
     
  5. Jun 25, 2010 #4

    Mark44

    Staff: Mentor

    f'(y) gives the slope at a point (y, f(y)) on the graph of f. If f'(y) > 0, the graph of f is increasing. If f'(y) < 0, the graph of f is decreasing.

    We know that f(1) = 0. Is the graph of f going up or down from there?

    BTW, this seems to be a calculus problem, so it should have been posted in the Calculus & Beyond section, not the Precalculus section.
     
  6. Jun 25, 2010 #5

    ehild

    User Avatar
    Homework Helper

    Write f(y) in terms of x=y-1≥0.

    ehild
     
  7. Jun 25, 2010 #6

    you can have f(y)<0 and f'(y)>=0 ,so we cannot get a contradiction
     
  8. Jun 25, 2010 #7
    If the graph is always increasing, and its value at the leftmost point (y=1) is 0, then can it be negative to the right (y>1)? You can also follow ehild's suggestion to get a more direct answer.
     
  9. Jun 25, 2010 #8
    Like this?

    [tex](x+1)^5+(x+1)^2-7(x+1)+5[/tex]
     
  10. Jun 25, 2010 #9
    Expand.
     
  11. Jun 25, 2010 #10

    Mark44

    Staff: Mentor

    This is really the long way around. You have the derivative -- f'(y) = 5y4 + 2y - y. It's a very simple matter to show that f'(y) >= 0 for y >= 1, hence the graph is increasing for y >= 1, and you're pretty much done.
     
  12. Jun 25, 2010 #11
    Factor the polynomial. Mathemtica gives:

    [tex]
    y^{5} + y^{2} -7y+5 = (y - 1)^{2} \, g_{3}(y)
    [/tex]

    where [itex]g_{3}(y)[/itex] is a polynomial of 3rd digree. What are the extremal values of the polynomial [itex]g_{3}(y)[/itex] in the interval [itex]y \ge 1[/itex]?
     
  13. Jun 25, 2010 #12

    ehild

    User Avatar
    Homework Helper

    It is not that difficult to expand, knowing the coefficients of (x+1)5 from Pascal's Triangle.

    (x+1)5=x5+5x4+10x3+10x2+5x+1

    ehild
     
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