What is the domain of $g(1 + x)$?

Question

$f(x)$ is a real valued function defined as

$$f(x) = 1 + x + (14x-3-8x^2)^{0.5}$$

$$g(x) = \big(f(x)\big)^{0.5}$$

What is the domain of $g(1 + x)$?

Answer 1

As I wrote in the comments, $$g(1+x)=\sqrt{2+x+\sqrt{3-2x-8x^2}}\;,\tag{1}$$ so the first step is to ensure that $3-2x-8x^2\ge 0$. This quadratic happens to factor: $3-2x-8x^2=(3+4x)(1-2x)$, so it’s equal to $0$ when $x=-\frac34$ and when $x=\frac12$. To see where it’s positive you can check to see where $3+4x$ and $1-2x$ have the same sign, but it’s easier to realize that the graph of $y=3-2x-8x^2$ is a parabola opening down, so it lies above the $x$-axis between the roots $-\frac34$ and $\frac12$. Thus, $3-2x-8x^2\ge 0$ when $-\frac34\le x\le\frac12$. (You must have made a sign error in the calculation in your comment.)

Okay, now we know that the domain is at most the interval $\left[-\frac34,\frac12\right]$, but there might be values of $x$ in that interval that make $$2+x+\sqrt{3-2x-8x^2}\tag{2}$$ negative. Are there?

At $x=-\frac34$, $(2)$ is equal to $2-\frac34=\frac54$, which is certainly not negative. As $x$ increases from $-\frac34$ to $\frac12$, $2+x$ also increases, so it won’t be a problem: it will always be at least $\frac54$. What about $\sqrt{3-2x-8x^2}$? On the interval $\left[-\frac34,\frac12\right]$ it’s always at least $0$, so it won’t be a problem either, and we can be sure that $(2)$ is always at least $\frac54$ when $-\frac34\le x\le\frac12$. In particular, it won’t be negative, so we can take its square root. Thus, the domain of $g(1+x)$ is the whole interval $\left[-\frac34,\frac12\right]$.