If f(x)=(x-3)^2(2x-3)^3 , then the point of inflexion is:

The correct answer is Option (2) → \( x=\displaystyle\frac{3}{2} \)

A point of inflexion is where f″(x) changes sign (the curve changes from concave to convex or back).

By the product rule, taking out the common factors:

    \( \displaystyle f'(x)=2(x-3)(2x-3)^3+6(x-3)^2(2x-3)^2 \)

    \( \displaystyle =2(x-3)(2x-3)^2\left[(2x-3)+3(x-3)\right] \)

    \( \displaystyle =2(2x-3)^2(x-3)(5x-12) \)

Writing \( g(x)=(x-3)(5x-12)=5x^2-27x+36 \), so that \( f'(x)=2(2x-3)^2\,g(x) \):

    \( \displaystyle f''(x)=2\left[4(2x-3)\,g(x)+(2x-3)^2(10x-27)\right] \)

    \( \displaystyle =2(2x-3)\left[4g(x)+(2x-3)(10x-27)\right] \)

    \( \displaystyle 4g(x)+(2x-3)(10x-27) \)

    \( \displaystyle =(20x^2-108x+144)+(20x^2-84x+81) \)

    \( \displaystyle =40x^2-192x+225 \)

    \( \displaystyle f''(x)=2(2x-3)(40x^2-192x+225) \)

At \( x=\frac{3}{2} \) the factor \( 2x-3 \) is 0 and changes sign, while \( 40x^2-192x+225=90-288+225=27\gt 0 \) there. So f″(x) changes sign at \( x=\frac{3}{2} \), which is a point of inflexion.

(The quadratic factor also gives two more points of inflexion, \( x=\frac{48\pm 3\sqrt{6}}{20} \), but they are not among the options.)

Why the other options are incorrect

• Option (1) \( x=3 \) — Incorrect. x = 3 is a critical point, f′(3) = 0, where f has a local minimum; f″(3) = 2(3)(9) = 54 ≠ 0.

• Option (3) \( x=\displaystyle\frac{12}{5} \) — Incorrect. \( x=\frac{12}{5} \) is a critical point, \( f'\left(\frac{12}{5}\right)=0 \), where f has a local maximum; \( f''\left(\frac{12}{5}\right)=2\left(\frac{9}{5}\right)\left(-\frac{27}{5}\right)=-\frac{486}{25}\ne 0 \).

• Option (4) \( x=0 \) — Incorrect. f″(0) = 2(−3)(225) = −1350 ≠ 0, so the concavity does not change at x = 0.

Tip: A factor raised to an odd power of 3 or more, like \( (2x-3)^3 \), gives a point of inflexion at its root; an even power, like \( (x-3)^2 \), gives a local extremum instead.