Step 1 :The derivative of the function does not exist where the function is not differentiable. For the absolute value function, this occurs where the argument of the absolute value function is zero.
Step 2 :So, we set the argument of the absolute value function equal to zero and solve for x:
Step 3 :\[\frac{1}{12} x^{2}-3=0\]
Step 4 :Multiplying through by 12 gives:
Step 5 :\[x^{2}-36=0\]
Step 6 :This factors to:
Step 7 :\[(x-6)(x+6)=0\]
Step 8 :So, the derivative does not exist at \(x=-6\) and \(x=6\).
Step 9 :For the graph, we know that the function is a parabola that opens upwards since the coefficient of \(x^{2}\) is positive. The vertex of the parabola is at \(x=0\), and the function is not differentiable at \(x=-6\) and \(x=6\), so the graph should have sharp turns at these points. Therefore, the correct graph is B.
Step 10 :For the relative minimum points, we know that a relative minimum occurs where the derivative changes from negative to positive. Since the function is not differentiable at \(x=-6\) and \(x=6\), these are the only possible points where the derivative could change sign. However, since the function is a parabola that opens upwards, the derivative is always positive except at these points, so there are no relative minimum points.
Step 11 :So, the final answers are \(x=-6\), \(x=6\), graph B, and there are no relative minimum points.