Find the inlervals on which the graph of I is concave upward, the intorvals on which tho graph off is concave downward, and the inflection points \[ f(x)=x^{3}-18 x^{2}+2 x+2 \] For what inlerval(s) of \( x \) is the graph of \( I \) concave upward? Select the correct duice bebw and, if necessary, fill in the answer box lo complete your chnice A. \( (6, \infty) \) (Type your answer in inleval notalion. Type an exacl answer. Uso a comna to separato answers as noodod.) B. The graph is never concave upward. Fot what inlerval(\$) of \( x \) is the graph of 1 concave downward? Select the conect choica below and, il necessary, fill in the answer hox to complete your choico. A. \( (-\infty, 6) \) (Type your answer in inlerval notation. Type an viacl answer. Use a comma to soparate answers as needed.) B. The graph is never concave downward. Delermine the \( x \), \( y \) coordinates of the inflection point of the graph of \( f \). Solect the correct choice below and, if necessary, fil in the answer box to complete your choice A. \( \square \) (Typs an ondered pair.) B. There are no inlloction ponts.

**Step 1. Find the second derivative.** Given \[ f(x)=x^3-18x^2+2x+2, \] first, compute the first derivative: \[ f'(x)=3x^2-36x+2. \] Then, compute the second derivative: \[ f''(x)=6x-36. \] **Step 2. Determine concavity.** - The graph is concave upward where \( f''(x)>0 \). Set up the inequality: \[ 6x-36>0. \] Dividing both sides by 6: \[ x-6>0 \quad \Longrightarrow \quad x>6. \] So, the graph is concave upward on \( (6,\infty) \). - The graph is concave downward where \( f''(x)<0 \). Set up the inequality: \[ 6x-36<0. \] Dividing both sides by 6: \[ x-6<0 \quad \Longrightarrow \quad x<6. \] So, the graph is concave downward on \( (-\infty,6) \). **Step 3. Find the inflection point.** An inflection point occurs where the concavity changes, i.e., where \( f''(x)=0 \). Set \( f''(x)=0 \): \[ 6x-36=0 \quad \Longrightarrow \quad x=6. \] Find the corresponding \( y \)-coordinate by evaluating \( f(6) \): \[ f(6)=6^3-18\cdot 6^2+2\cdot 6+2. \] Calculate step by step: \[ 6^3=216, \quad 6^2=36 \quad \Longrightarrow \quad 18\cdot 36=648, \] \[ f(6)=216-648+12+2= (216+12+2)-648 =230-648=-418. \] Thus, the inflection point is at \[ (6,-418). \] **Final Answers:** - The graph is concave upward on \( (6,\infty) \). - The graph is concave downward on \( (-\infty,6) \). - The inflection point is \((6,-418)\).