Determine the intervals for which the function is concave up or concave down. State the inflection points.
$f(x)=2x(x+9)^2$

• A. Concave down: $\(\left\)(-\(\infty\),\(\infty\]\right\))$ ; No Inflection Points
• B. Concave down: $\(\left\)(-\(\infty\),-6\(\right\))$; Concave up: $\(\left\)(-6,\(\infty\]\right\))$; Inflection pt: $\(\left\)(-6,-108\(\right\))$
• C. Concave down: $\(\left\)(-\(\infty\),6\(\right\))$; Concave up: $\(\left\)(6,\(\infty\]\right\))$; Inflection pt: $\(\left\)(6,2700\(\right\))$
• D. Concave down: $\(\left\)(-\(\infty\),0\(\right\))$; Concave up: $\(\left\)(0,\(\infty\]\right\))$; Inflection pt: $\(\left\)(0,0\(\right\))$

Answer: (B)

Verified step by step guidance

1. Step 1: To determine concavity, calculate the second derivative of the function f(x). Start by finding the first derivative f'(x) using the product rule and chain rule. The function is f(x) = 2x(x+9)^2.
2. Step 2: Differentiate f'(x) again to find the second derivative f''(x). Simplify the expression to make it easier to analyze the sign of f''(x).
3. Step 3: Set f''(x) = 0 to find the critical points where the concavity might change. Solve for x to determine these points.
4. Step 4: Use a test interval method or sign chart to analyze the sign of f''(x) in the intervals determined by the critical points. If f''(x) > 0, the function is concave up; if f''(x) < 0, the function is concave down.
5. Step 5: Identify the x-values where the concavity changes (from concave up to concave down or vice versa). These x-values are the inflection points. Substitute these x-values into the original function f(x) to find the corresponding y-coordinates of the inflection points.