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Thread: Quick range question: find range of f(x)=(x-5)/X^2-5x+6

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    Quick range question: find range of f(x)=(x-5)/X^2-5x+6

    Hi all

    quick question

    given f(x)=(x-5)/X^2-5x+6 find range

    EDIT

    ok so i think i got it, let me know if this seems right

    so write the equation as y=(x-5)/(x^2-5x+6)

    simplify y(x^2-5x+6)=(x-5)

    range of y will be values where discriminant > or equal to 0

    yx^2 - 5xy + 6y = x - 5
    yx^2 - 5xy +6y - x + 5 = 0

    quadratic is relative to x, where a = y, b = -(5y+1) and c = 6y+5

    discriminant = b^2 - 4ac so

    (-5y-1)^2 - 4y(6y+5)
    25y^2 +5y +5y +1 - 24y^2 - 20y
    y^2 -10y + 1 < or equal to 0
    y^2 - 10y + 1 = 0

    so range = (-infinty, 5-sqrt(24)] u [5+sqrt(24), infinity)

    basically I understand HOW questions like this work, i do not however understand WHY they work. any insight is appreciated

    EDIT

    http://www.analyzemath.com/DomainRan..._rational.html

    EDIT

    Looking for more than just this is how you find range of rational functions lol
    Last edited by frank789; 09-16-2017 at 07:03 PM.

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    Quote Originally Posted by frank789 View Post
    Hi all

    quick question

    given f(x)=(x-5)/X^2-5x+6 find range

    EDIT

    ok so i think i got it, let me know if this seems right

    so write the equation as y=(x-5)/(x^2-5x+6)

    simplify y(x^2-5x+6)=(x-5)

    range of y will be values where discriminant > or equal to 0

    yx^2 - 5xy + 6y = x - 5
    yx^2 - 5xy +6y - x + 5 = 0

    quadratic is relative to x, where a = y, b = -(5y+1) and c = 6y+5

    discriminant = b^2 - 4ac so

    (-5y-1)^2 - 4y(6y+5)
    25y^2 +5y +5y +1 - 24y^2 - 20y
    y^2 -10y + 1 < or equal to 0
    y^2 - 10y + 1 = 0

    so range = (-infinty, 5-sqrt(24)] u [5+sqrt(24), infinity)

    basically I understand HOW questions like this work, i do not however understand WHY they work. any insight is appreciated

    EDIT

    http://www.analyzemath.com/DomainRan..._rational.html

    EDIT

    Looking for more than just this is how you find range of rational functions lol
    y=(x-5)/(x^2-5x+6)

    y=(x-5)/(x^2 - 2x - 3x + 6)

    y=(x-5)/[(x - 2)*(x - 3)]

    Thus there are two vertical asymptotes at x = 2 & x = 3

    continue....
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  3. #3
    Elite Member mmm4444bot's Avatar
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    Quote Originally Posted by frank789 View Post
    f(x) = (x - 5)/(x^2 - 5x + 6)

    Grouping symbols around denominator are important.

    find range

    range of y will be values where discriminant > or equal to 0

    yx^2 - 5xy + 6y = x - 5

    yx^2 - 5xy +6y - x + 5 = 0

    discriminant = b^2 - 4ac

    (-5y - 1)^2 - 4y(6y + 5)

    y^2 - 10y + 1 < or equal to 0

    That less-than symbol ought to be a greater-than symbol.

    so range = (-infinity, 5 - sqrt(24)] ∪ [5 + sqrt(24), infinity)
    Your work looks good, and my answer matches yours.

    However, sqrt(24) can be simplified.


    … basically I understand HOW questions like this work, i do not however understand WHY they work. any insight is appreciated
    Can you be more specific? Are you asking why the Discriminant must be non-negative?

    PS: If you go on to study derivatives (in calculus), you will learn a more straight-forward method.
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

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    yeah I understand there are vertical asymptotes at 2 and 3 because they are restrictions on the denominator

    continuing on then, there must be a vertical asymptote at the restrictions on the inverse of the function.

    the discriminant of the equation will tell you the nature of the roots

    in this case, the sqrt(b^2 - 4ac) = 0 gives me 2 real solutions

    so i can find the solutions to be 5 + or - sqrt(24)

    i guess saying that the discriminant > 0 is my restriction so anything between these solutions would not be allowed since it would make the inequality false because it would return the sqrt of a negative

    i guess what im getting at is how do I tell the nature of my horizontal asymptotes from the discriminant of the inverse? I really want to avoid "you give me problem i give you an answer" and rather "you gave me this problem this is why it has this solution"

    im sorry if im not describing it well, please ask for any more info you need

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    Quote Originally Posted by mmm4444bot View Post
    Your work looks good, and my answer matches yours.

    However, sqrt(24) can be simplified.


    Can you be more specific? Are you asking why the Discriminant must be non-negative?

    PS: If you go on to study derivatives (in calculus), you will learn a more straight-forward method.
    unfourtunately, i have to do without calculus XD

    im more or less just asking, why can you pull the range from the discriminant? is it because that is the restriction on the function, i.e. values that are not allowed?

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    Quote Originally Posted by mmm4444bot View Post
    Your work looks good, and my answer matches yours.

    However, sqrt(24) can be simplified.


    Can you be more specific? Are you asking why the Discriminant must be non-negative?

    PS: If you go on to study derivatives (in calculus), you will learn a more straight-forward method.
    omg lol the discriminant must be non negative because it is the restriction on the function...thanks for talking me through it :P

  7. #7
    Elite Member mmm4444bot's Avatar
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    That's right. We want Real values for the range, so that Discriminant must be non-negative.
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

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