The function $f (x) = x^{3} - 6 x^{2} + 9 x + 1$ monotonically decreases for

1 < x < 3

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x < 3

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x > 1

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x > 3

x < 1

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Solution

The correct option is A $1 < x < 3$
$f (x) = x^{3} - 6 x^{2} + 9 x + 1$
$f (x) = 3 x^{2} - 12 x + 9$
For monotonically decreases
$f (x) < 0$
$3 x^{2} - 12 x + 9 < 0$
$x^{2} - 4 x + 3 < 0$
$(x - 3) (x - 1) < 0$
$[x ϵ (1, 3)]$

Suggest Corrections

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