Question

Show that the function \(f(x)=x^{3}-6x^{2}+12x,\ x\in R\), is an increasing function on \(R\).

Hint:

Differentiate f(x) and show f'(x) is a non-negative perfect square.

Answer 1

Step 1: Key Approach:
A function is increasing on \(R\) if \(f'(x)\ge0\) for all \(x\in R\).

Step 2: Differentiating:
\(f'(x)=3x^{2}-12x+12\).

Step 3: Rewriting as a perfect square:
\(f'(x)=3(x^{2}-4x+4)=3(x-2)^{2}\).

Final Answer:
Since \((x-2)^{2}\ge0\) for every real \(x\), \(f'(x)=3(x-2)^2\ge0\) for all \(x\in R\). Hence \(f\) is increasing on \(R\).\[ \boxed{f'(x)=3(x-2)^2\ge 0\ \Rightarrow f \text{ is increasing}} \]