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Thread: Math trouble! F(x) problem with a domain twist.

  1. #1
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    Math trouble! F(x) problem with a domain twist.

    Here is the problem:
    Let y = f(z) = √(4 − z^2) and z = g(x) = 2x + 3. Compute the compositiony = f(g(x)). Find the largest possible domain of x-values so that the composition y = f(g(x))is defined.


    The answer: −5/2 ≤ x ≤ −1/2

    Can someone guide me through the steps on how to reach such a conclusion? Thank you!
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  2. #2
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    Re: Math trouble! F(x) problem with a domain twist.

    I would write:

    $\displaystyle f(g(x))=\sqrt{4-(2x+3)^2}$

    Now, we require the radicand to be non-negative:

    $\displaystyle 4-(2x+3)^2\ge0$

    Factor the LHS:

    $\displaystyle (2+(2x+3))(2-(2x+3))\ge0$

    Distribute:

    $\displaystyle (2+2x+3)(2-2x-3)\ge0$

    Combine like terms:

    $\displaystyle (2x+5)(-2x-1)\ge0$

    Multiply by -1:

    $\displaystyle (2x+5)(2x+1)\le0$

    Since we have a quadratic on the left, whose graph is a parabola which opens up, we know it will be non-positive between and including the roots, which are:

    $\displaystyle x\in\left\{-\frac{5}{2},-\frac{1}{2}\right\}$

    Hence, the solution is given by:

    $\displaystyle -\frac{5}{2}\le x\le-\frac{1}{2}$

    Does that make sense?
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    Re: Math trouble! F(x) problem with a domain twist.

    Yes.. yes! Now it all makes perfect sense, I knew that I had a step missing. Thank you!
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