Find the inlorva's on which the graph of \( f \) is concave urwatd, the intervals \( f(x)=x^{3}-18 x^{2}+2 x+2 \) For what inlerval(s) of \( x \) is the graph of \( f \) concave upward? Select the cort A. \( \square \) B. The graph is nover concave upward.

To determine the intervals where the graph of \( f(x) = x^{3}-18x^{2}+2x+2 \) is concave upward, we need to find the second derivative, \( f''(x) \), and analyze where it is positive. First, let's find the first derivative: \[ f'(x) = 3x^{2} - 36x + 2. \] Next, we take the second derivative: \[ f''(x) = 6x - 36. \] Setting the second derivative equal to zero to find critical points: \[ 6x - 36 = 0 \Rightarrow x = 6. \] Now, we test intervals around \( x = 6 \): 1. For \( x < 6 \) (e.g., \( x = 0 \)): \( f''(0) = 6(0) - 36 = -36 < 0 \) (concave down). 2. For \( x > 6 \) (e.g., \( x = 7 \)): \( f''(7) = 6(7) - 36 = 6 > 0 \) (concave up). Thus, the graph is concave upward for \( x > 6 \). So the answer is: C. The graph is concave upward on the interval \( (6, \infty) \). To avoid confusion in interpreting this, it is also important to remember that the concept of concavity often arises in the context of optimization problems, where the second derivative informs us about the nature of critical points—whether they are local minima or maxima. This can be crucial in various fields from economics to engineering. Furthermore, understanding concavity can aid in estimating the behavior of functions, helping you to visualize where the graph bends. For instance, concave upward indicates that if you were to draw a line between any two points on the curve, that line would lie below the curve, hinting at potential minimum values in that region.