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Suppose the function f is defined as follows: f(x)=4x^2-5x+1

  1. Jun 29, 2015 #1
    1. The problem statement, all variables and given/known data
    Suppose the function f is defined as follows:
    f(x)=4x^2-5x+1
    Write down the D:f and solve the inequality f(x)≤0
    D:f (-∞,∞) because the function is a parabola, buy I am not sure how to solve the inequality, I would appreciate some help with this,

    Thank you


    2. Relevant equations
    f(x)=4x^2-5x+1
    f(x)≤0
    3. The attempt at a solution
    D:f (-∞,∞)
    4x^2-5x+1≤0

    Thank you
     
    Last edited: Jun 29, 2015
  2. jcsd
  3. Jun 29, 2015 #2
    Why don't you substitute for f(x) in the inequality and see what you get? If you don't know how to deal with a quadratic inequality, try finding the critical values and then think about the graph of the function for a bit.
     
  4. Jun 29, 2015 #3
    Thank you for your reply,
    Had a dumb moment there, I am busy working on it now.
     
  5. Jun 29, 2015 #4
    D:f (-∞,∞)
    4x^2-5x+1≤0
    (4x-1)(x-1)≤0
    4x-1=0
    4x=1
    x=(1/4)

    x-1=0
    x=1

    so x=1 and x=(1/4)
    1/4≤x≤1

    Thank you.

    Jaco
     
    Last edited: Jun 29, 2015
  6. Jun 29, 2015 #5

    RUber

    User Avatar
    Homework Helper

    You have found the points where the function equals zero. Your intervals for the function being less than zero will either be between those points or outside of those points.
    That is, your answer should be something like ##1/4 \leq x \leq 1 ## if between,
    ## x \leq 1/4 \text{ or } x \geq 1## if outside.
    One way to see what the proper interval is might be to test a few points.
     
    Last edited by a moderator: Jun 29, 2015
  7. Jun 29, 2015 #6
    Thank you Ruber,
    I have mended my answer above,
    Thank you again,

    Jaco
     
  8. Jun 29, 2015 #7

    RUber

    User Avatar
    Homework Helper

    I see you chose the correct interval, but did not express why it was correct. Did you plot a point inside or use a different method?
     
  9. Jun 29, 2015 #8
    Ruben,
    I substituted a few values into the equation see what was in fact true.
    Thank you,
     
  10. Jun 29, 2015 #9
    A good way to solve these problems is to look at the coefficient of the x^2 term. If it is positive, you'll get a U shaped graph, and any inequality of the type "less than or equal to" will give you a closed interval, whereas an inequality of the type "greater than or equal to" will give you two separate inequalities. The reverse is true if the x^2 coefficient is negative (an inverted U shape graph is produced). This is of course only valid if a solution to the equality between the quadratic function and whatever it is being compared with exists.
     
  11. Jun 29, 2015 #10

    Mark44

    Staff: Mentor

    Strictly speaking, you were working with an inequality, not an equation.
     
  12. Jun 29, 2015 #11

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    in addition to finding where [itex]4x^2- 5x+ 1[/itex] is equal to 0, you have determied that it can be factored as [itex](4x- 1)(x- 1)[/itex] so the inequality can be written as [itex](4x- 1)(x- 1)\le 0[/itex].

    Now, use the fact that "the product of two numbers is negative if and only if the two numbers are of different sign".
    Here, that tells you that either [itex]4x- 1\le 0[/itex] and [itex]x- 1\ge 0[/itex] or [itex]4x- 1\ge 0[/itex] and [itex]x- 1\le 0[/itex].
     
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