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

1 Answer
Jun 4, 2017

Answer:

Domain: #(-oo, oo)#
Range: #[-1/4, oo)#

Explanation:

Given:

#f(x) = x^2+x#

As with any polynomial, this is well defined for all values of #x#, so its domain is the whole of the Real numbers #RR#, in interval notation: #(-oo, oo)#

One way of finding the domain is to complete the square:

#x^2+x = (x+1/2)^2-1/4#

Note that:

#(x+1/2)^2 >= 0#

for any Real value of #x#, with equality when #x=-1/2#

So the minimum value of #f(x)# is:

#f(-1/2) = 0^2-1/4 = -1/4#

Since #f(x)# is continuous and unbounded, we can deduce that the range is #[-1/4, oo)#

One way of proving that goes as follows.

Let:

#y = x^2+x = (x+1/2)^2-1/4#

Add #1/4# to both sides to get:

#y+1/4 = (x+1/2)^2#

Transpose and take the square root of both sides, allowing for both positive and negative square roots to get:

#x+1/2 = +-sqrt(y+1/4)#

Subtract #1/2# from both sides to find:

#x = -1/2+-sqrt(y+1/4)#

So, provided #y in [-1/4, oo)#, there is at least one value of #x# such that #y = x^2+x#. That is: #y# is in the range of #f(x)#.