3. (12 points) Suppose $g, h$, and $j$ are differentiable functions with the values for the function and derivative given in the table: \begin{tabular}{|c|c|c|c|c|c|c|} \hline$x$ & $g(x)$ & $h(x)$ & $j(x)$ & $g^{\prime}(x)$ & $h^{\prime}(x)$ & $j^{\prime}(x)$ \\ \hline-1 & 3 & 0 & 1 & -1 & -2 & -2 \\ \hline 0 & 2 & 3 & 0 & -2 & 3 & -2 \\ \hline 1 & 0 & -1 & -2 & -2 & -2 & -1 \\ \hline \end{tabular} Let \[ f(x)=\frac{j(x)}{h(x)}+h(g(x)) \] Find $f^{\prime}(1)$.

Step 1 ：Given the function \(f(x)=\frac{j(x)}{h(x)}+h(g(x))\), we need to find \(f^{\prime}(1)\).

Step 2 ：To find \(f^{\prime}(1)\), we need to use the chain rule and quotient rule of differentiation.

Step 3 ：The chain rule states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function.

Step 4 ：The quotient rule states that the derivative of a quotient of two functions is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all over the square of the denominator.

Step 5 ：So, we first need to find the derivative of \(f(x)\), which is \(f^{\prime}(x) = \frac{j^{\prime}(x)h(x) - j(x)h^{\prime}(x)}{h^2(x)} + h^{\prime}(g(x))g^{\prime}(x)\).

Step 6 ：Then, we substitute \(x=1\) into the derivative function to find \(f^{\prime}(1)\).

Step 7 ：Given the values \(g(1) = 0\), \(h(1) = -1\), \(j(1) = -2\), \(g^{\prime}(1) = -2\), \(h^{\prime}(1) = -2\), and \(j^{\prime}(1) = -1\), we substitute these into the derivative function.

Step 8 ：Finally, we find that \(f^{\prime}(1) = \boxed{1.0}\).