If the function f(x) = kx3 − 9x2 + 9x + 3 is monotonically increasing in every interval, then

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k > 3 f\left( x ight) = k x^3 - 9 x^2 + 9x + 3 f'\left( x ight) = 3k x^2 - 18x + 9 = 3 \left( k x^2 - 6x + 3 ight) ext { Given:f(x) is monotonically increasing in every interval }. \Rightarrow f'\left( x ight) > 0 \Rightarrow 3 \left( k x^2 - 6x + 3 ight) > 0 \Rightarrow \left( k x^2 - 6x + 3 ight) > 0 \Rightarrow k > 0 ext { and } \left( - 6 ight)^2 - 4\left( k ight)\left( 3 ight) < 0 \left[ \because a x^2 + bx + c > 0 \Rightarrow a > 0 ext { and Disc} < 0 ight] \Rightarrow k > 0 ext { and } \left( - 6 ight)^2 - 4\left( k ight)\left( 3 ight) < 0 \Rightarrow k > 0 ext { and }36 - 12k < 0 \Rightarrow k > 0 ext { and }12k > 36 \Rightarrow k > 0 ext { and } k > 3 \Rightarrow k > 3

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