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

1 Answer
George C.
Aug 24, 2016

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]}

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