∵ 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≤0b2−4ac≤0⇒(−6)2−4(k)(3)≤0⇒36−12k≤012k≥36⇒k≥3 ∴ Solution is k≥3