Question

8. Show that the real valued function defined by f(x) = x 3 − 3x 2 + 3x + 1 increases for all x ∈ R

Answer 1

Let us show that the real valued function defined by $f(x) = x^3 − 3x^2 + 3x + 1$ increases for all $x ∈ \mathbb R.$ For this let us find the derivative: $f'(x) = 3x^2 − 6x + 3.$ Taking into account that $f'(x) = 3x^2 − 6x + 3= 3(x^2 − 2x + 1)=3(x-1)^2>0$ for all $x ∈ \mathbb R\setminus\{1\}$ and $f'(1)=0$, we conclude that the function $f$ increases for all $x ∈ \mathbb R.$