# What is the domain and range of f(x)=x^4-4x^3+4x^2+1?

May 20, 2015

I will assume that since the variable is called $x$, we are restricting ourselves to $x \in \mathbb{R}$. If so, $\mathbb{R}$ is the domain, since $f \left(x\right)$ is well defined for all $x \in \mathbb{R}$.

The highest order term is that in ${x}^{4}$, ensuring that:

$f \left(x\right) \to + \infty$ as $x \to - \infty$

and

$f \left(x\right) \to + \infty$ as $x \to + \infty$

The minimum value of $f \left(x\right)$ will occur at one of the zeros of the derivative:

$\frac{d}{\mathrm{dx}} f \left(x\right) = 4 {x}^{3} - 12 {x}^{2} + 8 x$

$= 4 x \left({x}^{2} - 3 x + 2\right)$

$= 4 x \left(x - 1\right) \left(x - 2\right)$

...that is when $x = 0$, $x = 1$ or $x = 2$.

Substituting these values of $x$ into the formula for $f \left(x\right)$, we find:

$f \left(0\right) = 1$, $f \left(1\right) = 2$ and $f \left(2\right) = 1$.

The quartic $f \left(x\right)$ is a sort of "W" shape with minimum value $1$.

So the range is $\left\{y \in \mathbb{R} : y \ge 1\right\}$