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

Aug 24, 2016

The domain of $f \left(x\right)$ is $\mathbb{R}$ and range is $\left[1 , \infty\right)$

#### Explanation:

Given:

$f \left(x\right) = {x}^{4} - 4 {x}^{3} + 4 {x}^{2} + 1$

I'm not sure what tools I'm really supposed to use under 'Algebra', but the easiest way to find where the turning points of $f \left(x\right)$ are is to differentiate it:

$f ' \left(x\right) = 4 {x}^{3} - 12 {x}^{2} + 8 x = 4 x \left({x}^{2} - 3 x + 2\right) = 4 x \left(x - 1\right) \left(x - 2\right)$

So $f ' \left(x\right) = 0$ when $x = 0$, $x = 1$, $x = 2$.

Since the coefficient of the highest degree term ${x}^{4}$ is positive, and all three turning points are Real and distinct, this is a classic "W" shaped quartic with minimum values at $x = 0$ and $x = 2$.

We find: $f \left(0\right) = 1$ and $f \left(2\right) = 16 - 32 + 16 + 1 = 1$

In common with any polynomial, the domain of $f \left(x\right)$ is the whole of the Real numbers $\mathbb{R}$ since the value is well defined for any Real value of $x$.

The range of $f \left(x\right)$ is $\left[1 , \infty\right)$ since $f \left(x\right)$ is continuous with minimum value $1$ and no upper bound.

graph{x^4-4x^3+4x^2+1 [-4, 6, -0.82, 4.18]}