If $f(x)=k x^3-9 x^2+9 x+3(k0)$ is increasing for all $x$, then

Question

If $f(x)=k x^3-9 x^2+9 x+3(k>0)$ is increasing for all $x$, then

Aicharya · Accepted Answer

$\because f(x)=k x^3-9 x^2+9 x+3(k>0)$ is increasing for all $x$. $\because f(x)$ is increasing, then $f^{\prime}(x) \geq 0$ $ \begin{aligned} & \Rightarrow \quad f^{\prime}(x) \geq 0 \ & \Rightarrow 3 k x^2-18 x+9 \geq 0 \ & \Rightarrow \quad k x^2-6 x+3 \geq 0 \ & \end{aligned} $ Using concept of quadratic polynomial for $k>0$ It is possible only, when $ \begin{aligned} & D \leq 0 \ & b^2-4 a c \leq 0 \ & \Rightarrow \quad(-6)^2-4(k)(3) \leq 0 \Rightarrow 36-12 k \leq 0 \ & \Rightarrow \quad 12 k \geq 36 \Rightarrow k \geq 3 \ & \end{aligned} $ $	herefore$ Solution is $k \geq 3$