I want to ask a question about the domain of two different functions in relation to an equation.
I was provided with two different functions:
$$f(x) = \frac{1}{x+1}, x > 3$$ $$f^{-1}(x) = \frac{1-x}{x}, 0 < x < \frac{1}{4}$$ $$g(x) = 2x^2 - 3, x \in \mathbb{R}$$
Now, I was asked to solve the equation:
$$fg(x) = \frac{1}{8}$$
This is what I did.
$$g(x) = f^{-1}(\frac{1}{8})$$ $$2x^2 - 3 = 7$$ $$2x^2 = 10$$ $$x^2 = 5$$ $$x = ±\sqrt{5}$$
This is what is given in the marking scheme, but I thought as $f(x)$ had domain $x>3$ I thought only $\sqrt{5}$ would be accepted.
Why are both values accepted?
Since we are dealing with the composite function $f\circ g$, the initial restriction (i.e. the domain of your equation) is: $$x\in D_g=\mathbb{R}, \\ g(x)\in D_f\Leftrightarrow g(x)\geq3\Leftrightarrow 2x^2-3\geq3\Leftrightarrow x^2\geq3$$ i.e.: $$x\in(-\infty,-\sqrt{3})\cup(\sqrt{3},+\infty)$$ This is why both solutions are accepted.