Find the inlervats of which tho praph of I is concowo upward, the inlorvals on which the graph of is concave downward, and tiw inflex.turn ponts \[ f(x)=x^{3}-18 x^{2}+2 x+2 \] For what inforvalis) of \( x \) is the graph of f concive upwnerd? Seinct the correct choice below and, if necossary It in the answor box lo complote your choice. A. \( (5, \infty) \) (Type your answer in inlarval notainn Type an eacl answer. Uso a comma to soparato antowors as noodod) B. The grap) is nover concave upwand. For whal iniorval(s) ol \( x \) is tho graph of f concave downward? Selod the cortost choico bolow and, it necoseary fill in the answer boxtlo cormplote yeur chaico. A. \( (-\infty \). 6\( ) \) (Typo yout answor in inlerval nodation Type an cxact answer: Uso a comma lo separato answors as nosdod.) B. The g'aph is newne crncave downard A. \( \square \) (Typo en ordoted prair) B. Then nin no enlection poants

To find the intervals of concavity, we first need to compute the second derivative of the function \(f(x) = x^3 - 18x^2 + 2x + 2\). The first derivative is \(f'(x) = 3x^2 - 36x + 2\), and the second derivative is \(f''(x) = 6x - 36\). Setting \(f''(x) = 0\) gives us \(x = 6\). Now examining the intervals: - For \(x < 6\), \(f''(x)\) is negative, indicating the graph is concave downward in the interval \((-∞, 6)\). - For \(x > 6\), \(f''(x)\) is positive, meaning the graph is concave upward in the interval \((6, ∞)\). Thus, the inflection point occurs at \(x = 6\), where the concavity changes. The final answers for the intervals of concavity are: - For concave upward: \((6, \infty)\) - For concave downward: \((-\infty, 6)\) - Inflection point: \(x = 6\) You've got this! It’s just like a roller coaster: at each change of direction, the fun transforms from soaring up high to diving low! Enjoy the curves!