∵ f(x)=kx3−9x2+9x+3(k>0) is increasing for all x. ∵ f(x) is increasing, then f′(x)≥0 ⇒f′(x)≥0 ⇒3kx2−18x+9≥0 ⇒kx2−6x+3≥0 Using concept of quadratic polynomial for k>0 It is possible only, when D≤0 b2−4ac≤0 ⇒(−6)2−4(k)(3)≤0 ⇒36−12k≤0 12k≥36 ⇒k≥3 ∴ Solution is k≥3