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Math Help - Finding range of function by using algebra?

  1. #1
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    Finding range of function by using algebra?

    Use algebrad the range of f(x) = 3x^2+10/x^2-9
    i remember our teacher told us to solve for A, any real number so I did
    3x^2+10/x^2-9 =A
    3x^2+10=A(x^2-9)
    3x^2+10=Ax^2-9A
    9A+10=Ax^2-3x^2
    9A+10=x^2(A-3)
    9A+10/A-3 = x^2
    A cannot equal 3
    I don't know what the next step is
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  2. #2
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    Quote Originally Posted by dorkymichelle View Post
    Use algebrad the range of \bold{\color{red}f(x) = (3x^2+10)/(x^2-9)}
    i remember our teacher told us to solve for A, any real number so I did
    3x^2+10/x^2-9 =A
    3x^2+10=A(x^2-9)
    3x^2+10=Ax^2-9A
    9A+10=Ax^2-3x^2
    9A+10=x^2(A-3)
    9A+10/A-3 = x^2
    A cannot equal 3
    I don't know what the next step is
    1. Since x^2 \geq 0 for all x \in \mathbb{R} the LHS of the equation must be greater zero too.

    2. A quotient is greater (or equal) zero if the signs of denominator and numerator are equal:

    9A+10\geq 0 \wedge A-3>0~\vee~9A+10\leq 0\wedge A-3<0

    A\geq -\frac{10}9 \wedge A>3~\vee~A\leq -\frac{10}9\wedge A<3

    \boxed{A>3~\vee~A\leq -\frac{10}9}

    That's the range of the function.

    Btw: Please write numerator and denominator of a quotient in brackets. Otherwise you mean: f(x) = 3x^2+\frac{10}{x^2}-9
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  3. #3
    MHF Contributor red_dog's Avatar
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    x^2=\frac{9A+10}{A-3}

    The equation must have real solutions. Then the condition is that

    \frac{9A+10}{A-3}\geq 0\Rightarrow A\in\left(\left.\-\infty,-\frac{10}{9}\right]\right.\cup(3,\infty)
    Last edited by red_dog; December 11th 2009 at 11:58 PM.
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  4. #4
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    I understand the part where it has to be greater than 3, because that would make the demoniator zero. But i don't understand why there is a restriction on -10/9
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  5. #5
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by dorkymichelle View Post
    I understand the part where it has to be greater than 3, because that would make the demoniator zero. But i don't understand why there is a restriction on -10/9
    Think of it this way. You found your critical points to be 3 and -10/9 by setting the numerator and denominator to 0

    Now test a value from - infinity to -10/9, next a value from -10/9 to 3, and then from 3 to positive infinity.

    You will notice that end up with a positive value in the intervals of (- infinity to -10/9) and (3 to positive infinity).

    You will end up with a negative value in the interval of (-10/9 to 3)

    Now since you can only have real solutions you can't use the negative interval.

    The reason why you can't use the negative interval is when you take a square root of a negative number you get an imagnery number.

    Does this make sense?
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  6. #6
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    So basicly, 9A+10/A-3 cannot be a negative number because you can't square root a negative number and still have a real number, because of x^2 and -10/9 is the number that makes the fraction 0 and any number less than that would make the fraction negative correct?
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  7. #7
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by dorkymichelle View Post
    So basicly, 9A+10/A-3 cannot be a negative number because you can't square root a negative number and still have a real number, because of x^2 and -10/9 is the number that makes the fraction 0 and any number less than that would make the fraction negative correct?
    Close but not quite.

    Yes

    \frac{9A+10}{A-3}

    has to be positive but the solution set is like red dog said. Notice how that includes values less than \frac{-10}{9}
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