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Question 8.7.2: Find possible inflection points for f (x) = 1/9x³ − 1/6x² − ......

Find possible inflection points for

f(x)={\textstyle\frac{1}{9}}x^{3}-{\textstyle\frac{1}{6}}x^{2}-{\textstyle\frac{2}{3}}x+1.
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From Example 8.6.1, we have f^{\prime\prime}(x) =\frac{ 2}{3}x − \frac{1}{3}= \frac{2}{3}(x − \frac{1}{2}). Hence, f^{\prime\prime}(x) ≤ 0 for x ≤ 1/2, whereas f^{\prime\prime}(1/2) = 0 and f^{\prime\prime}(x) ≥ 0 for x > 1/2. According to part (ii) in Theorem 8.7.1, x = 1/2 is an inflection point for f .

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