Question

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

Answer 1

The domain of `f(x)` is `RR` and range is `[1, oo)`

Explanation:

Given:

`f(x) = x^4-4x^3+4x^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(x)` are is to differentiate it:

`f'(x) = 4x^3-12x^2+8x = 4x(x^2-3x+2) = 4x(x-1)(x-2)`

So `f'(x) = 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(0) = 1` and `f(2) = 16-32+16+1 = 1`

In common with any polynomial, the domain of `f(x)` is the whole of the Real numbers `RR` since the value is well defined for any Real value of `x`.

The range of `f(x)` is `[1, oo)` since `f(x)` is continuous with minimum value `1` and no upper bound.

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