6 mark for review if $f(x)=x(x + 1)(x - 2)^2$, then the graph of $f$ has inflection points when $x=$ a -1 only b 2 only c -1 and 0 only d -1 and 2 only e -1, 0, and 2 only

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# Explanation: ## Step1: Recall Inflection Point Condition An inflection point of $ f(x) $ occurs where $ f''(x) = 0 $ and $ f''(x) $ changes sign (or the concavity changes). First, we need to analyze $ f''(x)=x(x + 1)(x - 2)^2 $. ## Step2: Find Critical Points of $ f''(x) $ Set $ f''(x)=0 $. Solve $ x(x + 1)(x - 2)^2=0 $. The roots are $ x = 0 $, $ x=-1 $, and $ x = 2 $ (with $ x = 2 $ being a double root). ## Step3: Analyze Sign Changes of $ f''(x) $ - For $ x<-1 $, let's pick $ x=-2 $: $ (-2)(-2 + 1)(-2 - 2)^2=(-2)(-1)(16)=32>0 $. - For $ -10 $. So $ f''(x) $ changes sign at $ x = 0 $. - For $ x>2 $, pick $ x = 3 $: $ (3)(3 + 1)(3 - 2)^2=(3)(4)(1)=12>0 $. For $ x = 2 $, the factor $ (x - 2)^2 $ is always non - negative, and the sign of $ f''(x) $ around $ x = 2 $ (left: $ 00 $; right: $ x>2 $, $ f''(x)>0 $) does not change. So $ x = 2 $ is not an inflection point. # Answer: C. -1 and 0 only

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Explanation:

Step1: Recall Inflection Point Condition

Step2: Find Critical Points of \( f''(x) \)

Step3: Analyze Sign Changes of \( f''(x) \)

Answer: