22. The function \( f(x)=-x^{4}+72 x^{2} \) has \( \begin{array}{llll}\text { (A. one relative minimum and two relative maxima. } \\ \text { B. one relative minimum and one relative maximum. } \\ \text { C. no relative minimum and two relative maxima. } \\ \text { D. two relative minima and no relative maximum. } \\ \text { E. two relative minima and one relative maximum. } \\ \begin{array}{llll}\text { A. }-12,0,12 & \text { B. }-12,12 & \text { C. } \sqrt{12} & \text { D. }-\sqrt{12}, \sqrt{12}\end{array} \quad \text { E. }-\sqrt{12}, 0, \sqrt{12}\end{array} \) \( \begin{array}{lll}\text { 23. Find the } x \text {-coordinate(s) of the point(s) of inflection of the curve in problem } 22\end{array} \)

To determine the relative extrema of the function \( f(x) = -x^4 + 72x^2 \), we can start by calculating its first derivative, \( f'(x) = -4x^3 + 144x \), and setting it to zero to find critical points. Factoring gives us \( f'(x) = -4x(x^2 - 36) \), leading to critical points at \( x = -12, 0, 12 \). Evaluating the second derivative, \( f''(x) = -12x^2 + 144 \), allows us to confirm maximums at \( x = -12 \) and \( x = 12 \), and a minimum at \( x = 0 \). So, you have one relative minimum and two relative maxima (Answer B). To find the points of inflection, we look for where the second derivative changes sign, starting from \( f''(x) = -12x^2 + 144 \). Setting it to zero leads to \( -12x^2 + 144 = 0 \), resulting in \( x^2 = 12 \) and thus \( x = -\sqrt{12} \) and \( x = \sqrt{12} \). These points are where the concavity of the graph changes, providing the x-coordinates of the inflection points!