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Math Help - Solving range algebraically

  1. #1
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    Question Solving range algebraically

    Let's say you have a function in which you have a linear function divided by a parabola. How do you find the range of this? Also, how do you find the range when you have something such as a parabola divided by a parabola? Thanks for your help.
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  2. #2
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    Re: Solving range algebraically

    Hey symphonicity.

    The answer will depend on the nature of the coefficients, but as a guide you should find out:

    a) When the denominator becomes zero (if it becomes zero)
    b) What the limit of the function is at plus or minus infinity
    c) When the first derivatives are zero (and what kind of points they are)

    These should give the answers to get the range.
    Thanks from MarkFL
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  3. #3
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    Re: Solving range algebraically

    Quote Originally Posted by symphonicity View Post
    Let's say you have a function in which you have a linear function divided by a parabola. How do you find the range of this? Also, how do you find the range when you have something such as a parabola divided by a parabola? Thanks for your help.
    A nice method is to find the domain of the inverse relation, if it's possible to write the inverse relation as y in terms of x...
    Thanks from MarkFL
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  4. #4
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    Re: Solving range algebraically

    The "range" of function y= f(x) is the set of all values of y given by some value of x in the domain. What you can do is look for the x that will give a value of y. And that is equivalent to finding the inverse function as Prove It suggested.

    For example, to find the range of f(x)= x^2- 3x+ 4, we can write y= x^2- 3x+ 4 and attempt to solve for x. Since this particular function is quadratic, x^2- 3x+ (4-y)= 0, the quadratic formula works nicely: x= \frac{3\pm\sqrt{9- 4(4- y)}}{2}= \frac{3\pm\sqrt{4y- 7}}{2}. There will exist real x satisfying that as long as 4y- 7\ge 0 so the range is "all y greater than or equal to 7/4".
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