In which interval(s) is the graph of the following function concave up? \[ f(x)=\frac{x^{4}}{4}-x^{3}-\frac{9 x^{2}}{2}+7 \] a. $x \in(-3,1)$ b. $x \in(-1,3)$ c. $x \in(-\infty,-1)$ and $x \in(3, \infty)$ d. $x \in(-\infty,-3)$ and $x \in(1, \infty)$

Step 1 ：Given the function \(f(x)=\frac{x^{4}}{4}-x^{3}-\frac{9 x^{2}}{2}+7\), we need to find the intervals where the function is concave up.

Step 2 ：To do this, we first find the second derivative of the function. The second derivative of \(f(x)\) is \(f''(x) = 3x^{2} - 6x - 9\).

Step 3 ：We then set the second derivative greater than zero to find the intervals where the function is concave up. This gives us the inequality \(3x^{2} - 6x - 9 > 0\).

Step 4 ：Solving this inequality, we find that the function is concave up for \(x \in(-\infty,-1)\) and \(x \in(3, \infty)\).

Step 5 ：\(\boxed{\text{Final Answer: The correct answer is c. } x \in(-\infty,-1)\text{ and }x \in(3, \infty)}\)