How do you find the domain and range of #f(x)=x(x+2)(x-2)(x+4)#?

1 Answer
Nov 26, 2017


See explanation. #-oo < x < oo, -16 <= y < oo#


This function is a polynomial. As a polynomial with no apparent discontinuities (i.e. not being divided by any sort of function other than the number 1), its domain will be all real numbers. #-oo < x < oo#

For the range, however, we must actually look at the function.

#f(x) = x(x+2)(x-2)(x+4)#

The function has 3 binomials of the form #x+c_1# being multiplied together, and that product being multiplied by #x#. This means that our highest order term will be of the form #x^4#, which is always positive. This means that as #x# increases towards positive infinity (or decreases towards negative infinity), the #x^4# term will approach positive infinity, and will outweigh every other term in the expression.

What this means for our purposes, is that there is an absolute minimum value to the function, specifically because end behavior trends towards infinity, and because the function is continuous throughout the domain of all real numbers.

For this purpose, it behooves us to use a graphing calculator, such as the TI-83+, to graph the function. Once we have done this, the calculator will present us with the option to calculate the minimum. Doing this, we find that the minimum in the function occurs at #(x_min, y_min) = approx (-3.236, -16)#.

If this is the absolute minimum, and the function trends towards positive infinity as x approaches positive or negative infinity, our range is #-16<= y < oo#