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Find a point on curve of f(x) = x(x - 3)2 when tangent is parallel to equation y = 0 in [0, 3]

Find a point on curve of f(x) = x(x - 3)2 when tangent is parallel to equation y = 0 in [0, 3] Correct Answer (1, 4)

Given:

Y = f(x) = x(x - 3)2 = x3 – 6x2 + 9x ⇒ Polynomial

F(x) is

i) Continuous on

And ii) differentiable on (0, 3)

iii) f(0) = f(3) = 0

Thus, all three conditions of Rolle’s Theorem are satisfied.

By Rolle’s Theorem, C ϵ (0, 3) such that f’(c) = 0

f(C) = C (C - 3)2

f(C) = C

f(C) = C3 – 6c2 + 9C

Differentiating,

f’(C) = 3C2 – 12C + 9 = 0

∴ C = 1, 3

But, C = 3 ∉ (0, 3)

∴ C = 1

If x = C = 1, then y (1) = f (1) = 1 (1 - 3)2 = 4

∴ At a point (x, y) = (1, 4) on given curve, the tangent is parallel to x-axis.

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