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

Step 1 ：We are given the function $f(x)=3x^2-5$ and $(f\circ g)(x)=1-3x$, 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 $x\le 2$.

Step 6 ：For $x\le 2$, $\sqrt{2-x}$ 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 ：$\boxed{{\displaystyle g(x)=-\sqrt{2-x}}}$ and $\boxed{{\displaystyle g(x)=\sqrt{2-x}}}$ are the final answers.