Given $f(x)=3 x^{2}-5 \quad$ and $(f \circ g)(x)=1-3 x, \quad$ find $g(x)$

Step 1 ：We are given the function \(f(x)=3 x^{2}-5\) and \((f \circ g)(x)=1-3 x\), and we are asked to find \(g(x)\).

Step 2 ：We know that \((f \circ g)(x)\) means \(f(g(x))\). So, we can substitute \(g(x)\) into \(f(x)\) and set it equal to \(1-3x\). Then, we can solve for \(g(x)\).

Step 3 ：Setting \(3g^{2}-5\) equal to \(1-3x\), we get the equation \(3g^{2}-5 = 1-3x\).

Step 4 ：Solving this equation, we get two solutions for \(g(x)\): \(-\sqrt{2 - x}\) and \(\sqrt{2 - x}\).

Step 5 ：However, we need to consider the domain of the function \(g(x)\). Since \(g(x)\) is inside the square function in \(f(g(x))\), it must be real. Therefore, the domain of \(g(x)\) is \(x \leq 2\).

Step 6 ：For \(x \leq 2\), \(\sqrt{2 - x}\) is real and non-negative, while \(-\sqrt{2 - x}\) is non-positive. Therefore, both \(-\sqrt{2 - x}\) and \(\sqrt{2 - x}\) are valid solutions for \(g(x)\).

Step 7 ：\(\boxed{g(x) = -\sqrt{2 - x}}\) and \(\boxed{g(x) = \sqrt{2 - x}}\) are the final answers.