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

We begin with the function \[ f(x)=x^3-18x^2+2x+2. \] **Step 1. Compute the Second Derivative** The second derivative \( f''(x) \) tells us about the concavity. First, compute the first derivative: \[ f'(x)=3x^2-36x+2. \] Now, differentiate \( f'(x) \) to obtain the second derivative: \[ f''(x)=6x-36. \] --- **Step 2. Determine the Inflexion Point** An inflection point occurs where the concavity changes, which is where \( f''(x)=0 \): \[ 6x-36=0 \quad \Longrightarrow \quad x=6. \] To find the corresponding \( y \)-coordinate, substitute \( x=6 \) into \( f(x) \): \[ f(6)=6^3-18\cdot6^2+2\cdot6+2=216-648+12+2. \] Simplify step by step: \[ 216-648=-432, \quad -432+12=-420, \quad -420+2=-418. \] Thus, the inflection point is at \[ (6, -418). \] --- **Step 3. Find the Intervals of Concavity** 1. **Concave Upward:** The graph is concave upward where \( f''(x)>0 \). \[ 6x-36>0 \quad \Longrightarrow \quad 6x>36 \quad \Longrightarrow \quad x>6. \] So, the graph is concave upward on the interval: \[ (6, \infty). \] 2. **Concave Downward:** The graph is concave downward where \( f''(x)<0 \). \[ 6x-36<0 \quad \Longrightarrow \quad 6x<36 \quad \Longrightarrow \quad x<6. \] Thus, the graph is concave downward on the interval: \[ (-\infty, 6). \] --- **Final Answers in the Given Format:** - For what intervals of \( x \) is the graph of \( f \) concave upward? The correct answer is not \( (5, \infty) \); the graph is concave upward on \( (6, \infty) \). - For what intervals of \( x \) is the graph of \( f \) concave downward? The correct answer is \( (-\infty, 6) \). The function has an inflection point at \( (6, -418) \).