which function has the same range as f(x)=-2\\sqrt{x - 3}+8?\ng(x)=\\sqrt{x - 3}-8\ng(x)=\\sqrt{x…

Explanation:

Step1: Analyze the range of $f(x)=-2\sqrt{x - 3}+8$

The square - root function $\sqrt{x-3}$ has a domain $x\geq3$, and $\sqrt{x - 3}\geq0$. Then $-2\sqrt{x - 3}\leq0$, and $f(x)=-2\sqrt{x - 3}+8\leq8$. So the range of $f(x)$ is $(-\infty,8]$.

Step2: Analyze the range of each $g(x)$ function

For $g(x)=\sqrt{x - 3}-8$

Since $\sqrt{x - 3}\geq0$, then $g(x)=\sqrt{x - 3}-8\geq - 8$, range is $[-8,\infty)$.

For $g(x)=\sqrt{x - 3}+8$

Since $\sqrt{x - 3}\geq0$, then $g(x)=\sqrt{x - 3}+8\geq8$, range is $[8,\infty)$.

For $g(x)=-\sqrt{x + 3}+8$

The domain is $x\geq - 3$, and $\sqrt{x + 3}\geq0$, so $-\sqrt{x + 3}\leq0$, and $g(x)=-\sqrt{x + 3}+8\leq8$, range is $(-\infty,8]$.

For $g(x)=-\sqrt{x - 3}-8$

Since $\sqrt{x - 3}\geq0$, then $-\sqrt{x - 3}\leq0$, and $g(x)=-\sqrt{x - 3}-8\leq - 8$, range is $(-\infty,-8]$.

Answer:

$g(x)=-\sqrt{x + 3}+8$