Maximize $f(x, y)=4 x+y$ subject to $x^{2}+y=3$ without using the method of Lagrange multipliers; instead, solve the constraint for $x$ or $y$ and substitute into $f(x, y)$. Use the constraint to rewrite $f(x, y)=4 x+y$ as a function of one variable, $g(x)$. \[ g(x)= \] The maximum value of $f(x, y)=4 x+y$ subject to $x^{2}+y=3$ occurs at the point (Type an ordered pair.) The maximum value of $f(x, y)=4 x+y$ subject to $x^{2}+y=3$ is (Simplify your answer.)

Step 1 ：First, solve the constraint equation for y: \(y = 3 - x^2\).

Step 2 ：Substitute y into the function f(x, y) to get a function of one variable, g(x): \(g(x) = 4x + 3 - x^2\).

Step 3 ：Find the maximum of g(x) by taking the derivative and setting it equal to zero: \(g'(x) = 4 - 2x = 0\).

Step 4 ：Solve for x to find the critical points: \(x = 2\).

Step 5 ：Substitute x = 2 into the function g(x) to find the maximum value: \(g(2) = 7\).

Step 6 ：Substitute x = 2 into the constraint equation to find the corresponding y value: \(y = 3 - 2^2 = -1\).

Step 7 ：Final Answer: The maximum value of \(f(x, y)=4 x+y\) subject to \(x^{2}+y=3\) occurs at the point \(\boxed{(2, -1)}\). The maximum value of \(f(x, y)=4 x+y\) subject to \(x^{2}+y=3\) is \(\boxed{7}\).