How do you find the domain and range of # P(x)=2x(x+1)(x-2)#?

2 Answers
Feb 14, 2018

Answer:

The domain and range are both all real numbers.

Explanation:

In this polynomial,

# P(x)=2x(x+1)(x-2)#,

the domain is all real numbers because any #x# value plugged in will yield a valid #y# value. (Note: this is true for all polynomials.)

The range is a bit trickier to figure out. If you were to expand this polynomial out, you would get something like

#P(x)=?x^3+?x^2 +?x+?#

where the question marks represent an unknown number. Since the degree of this polynomial is #3#, that means it is classified as a cubic. Cubics can look like this:

graph{(x-1)^3+2(x-1)^2-sqrt(2)/2 [-7, 7, -5, 5]}

Or this:

graph{-x^3+2(x)^2 [-7, 7, -5, 5]}

The one thing all cubics have in common is the end behavior, which is what direction they end in.

All the graphs start at the top and end at the bottom (or vice versa), which means that at some point, they must cross every single #y#-value. Therefore, their ranges are all real numbers.

That is also why #P(x)#'s range is all real numbers.

Feb 14, 2018

Answer:

Domain: #(-oo, oo)#
Range: #(-oo, oo)#

Explanation:

If you multiply all of those factors together, you get a third-degree polynomial with a positive leading coefficient. Base on this information, we can conclude that the end behavior is as follows:

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

Therefore, since all polynomials are continuous, the range #f(x)# is #(-oo, oo)#

Polynomials also have no restrictions on domain, so the domain #x# is also #(-oo, oo)#