Answer the following questions for the function \[ f(x)=x \sqrt{x^{2}+9} \] defined on the interval $-7 \leq x \leq 5$. $f(x)$ is concave down on the interval $x=$ to $x=$ $f(x)$ is concave up on the interval $x=$ to $x=$ The inflection point for this function is at $x=$ The minimum for this function occurs at $x=$ The maximum for this function occurs at $x=$

Step 1 ：First, we need to find the derivative of the function. The derivative of \(f(x) = x \sqrt{x^2 + 9}\) is \(f'(x) = \sqrt{x^2 + 9} + \frac{x^2}{\sqrt{x^2 + 9}}\).

Step 2 ：Next, we find the second derivative of the function. The second derivative of \(f(x)\) is \(f''(x) = \frac{x}{\sqrt{x^2 + 9}} - \frac{x^3}{(x^2 + 9)^{3/2}}\).

Step 3 ：Setting \(f''(x) = 0\), we get \(x = 0\) and \(x = \pm 3\).

Step 4 ：By checking the sign of \(f''(x)\) in each interval, we find that \(f(x)\) is concave down on the interval \(x = -7\) to \(x = -3\) and \(x = 3\) to \(x = 5\).

Step 5 ：Similarly, \(f(x)\) is concave up on the interval \(x = -3\) to \(x = 3\).

Step 6 ：The inflection point for this function is at \(x = \pm 3\).

Step 7 ：To find the minimum and maximum of the function, we need to check the critical points and endpoints. The critical points are where \(f'(x) = 0\), which gives us \(x = 0\).

Step 8 ：By substituting \(x = -7\), \(x = 0\), and \(x = 5\) into \(f(x)\), we find that the minimum for this function occurs at \(x = -7\) and the maximum for this function occurs at \(x = 5\).

Step 9 ：Thus, the function \(f(x) = x \sqrt{x^2 + 9}\) is concave down on the interval \(x = -7\) to \(x = -3\) and \(x = 3\) to \(x = 5\), concave up on the interval \(x = -3\) to \(x = 3\), the inflection point is at \(x = \pm 3\), the minimum occurs at \(x = -7\), and the maximum occurs at \(x = 5\).

Step 10 ：Finally, we check our results and find that they meet the requirements of the problem. So, the final answer is \(\boxed{-7}\) for the minimum and \(\boxed{5}\) for the maximum.