Find the point of inflection for f(x) = \(x^4\over 12\) – \(5x^3\over 6\) + \(3x^2\) + 7.

Solution :

f(x) = \(x^4\over 12\) – \(5x^3\over 6\) + \(3x^2\) + 7.

f'(x) = \(x^3\over 3\) – \(5x^2\over 2\) + 6x

f”(x) = \(x^2\) – 5x + 6

Since, f”(x) = 0 at point of inflection.

\(\implies\) \(x^2\) – 5x + 6 = 0

\(\implies\) x = 2 and x = 3

Hence, points of inflection are 2 and 3.

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