Range and injectivity

Text

Given $f(x)=x^{2}-4x+1$ defined on $\mathbb{R}$, find the range and decide whether it is injective.

Solution

The minimum is at the vertex: $x_V=-\dfrac{-4}{2}=2$, $f(2)=4-8+1=-3$. Since the parabola opens upward, the range is $[-3,+\infty)$. It is not injective on $\mathbb{R}$: for example $f(0)=f(4)=1$. $\boxed{\,\text{Im}=[-3,+\infty),\ \text{not injective}\,}$