Q.267. Point of inflection.

Question 267.
Let f (x) = x (2x - 15)^(2/3), for all x. The graph of f has
(a) no points of inflextion at all      (b) a point of inflection at x = 9
(c) a point of inflection at x = 7      (d) a point of inflection at x = 8
(e) a point of inflection at x = 6.

Answer 267.
The function y = f (x) has a point of inflection at a value of x for which
f "(x) = 0 and f "'(x) ≠ 0.

f (x) = x * (2x - 15)^(2/3)

=> f '(x)
= (2x - 15)^(2/3) + (4x/3) (2x - 15)^(-1/3)
= (2x - 15 + 4x/3) * (2x - 15)^(-1/3)
= (5/3) (2x - 9) * (2x - 15)^(-1/3)
 => f "(x)
= (10/3) (2x - 15)^(-1/3) - (10/3) (2x - 9) * (2x - 15)^(-4/3)
= [(10/3)(2x - 15) - (10/9)(2x - 9)] * (2x - 15)^(-4/3)
= (40/9) (x - 9) * (2x - 15)^(-4/3)
 => f "'(x)
= (40/9) * (2x - 15)^(-4/3) - (320/27) (x - 9) * (2x - 15)^(-7/3)
= (40/27) (6x - 45 - 8x + 72) * (2x - 15)^(-7/3)
= (40/27) (27 - 2x) * (2x - 15)^(-7/3)

f (x) = 0 for x = 0 or 7.5,   f '(x) = 0 for x = 4.5
f "(0) = 0 for x = 9  and  f"'(9) ≠ 0.

=> there is a point of inflection at x = 9

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