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

1. ## 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!  2. ## 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?

3. ## 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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