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

1 Answer
May 20, 2015

I will assume that since the variable is called x, we are restricting ourselves to x in RR. If so, RR is the domain, since f(x) is well defined for all x in RR.

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

f(x) -> +oo as x -> -oo

and

f(x)->+oo as x -> +oo

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

d/(dx) f(x) = 4x^3-12x^2+8x

= 4x(x^2-3x+2)

= 4x(x-1)(x-2)

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

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

f(0) = 1, f(1) = 2 and f(2) = 1.

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

So the range is { y in RR: y >= 1 }