What is the domain and range of #f(x) = x^2 + 4x – 6#?

1 Answer
Alan P.
Oct 1, 2015

Domain: #RR#
Range: #RR >= -10#

Explanation:

#f(x)=x^2+4x-6#
is valid for all Real values of #x#
and therefore the Domain is all Real values i.e. #RR#

To determine the Range, we need to find what values of #f(x)# can be generated by this function.
Probably the simplest way to do this is to generate the inverse relation. For this I will use #y# in place of #f(x)# (just because I find it easier to work with).

#y=x^2+4x-6#

Reversing the sides and completing the square:
#color(white)("XXX")(x^2+4x+4) - 10 = y#

Re-writing as a square and adding #10# to both sides:
#color(white)("XXX")(x+2)^2=y+10#

Taking the square root of both sides
#color(white)("XXX")x+2 = +-sqrt(y+10)#

Subtracting #2# from both sides
#color(white)("XXX")x= +-sqrt(y+10) -2#

Assuming that we are restricted to Real values (i.e. non-Complex), this expression is valid provided:
#color(white)("XXX")y>=-10#
#color(white)("XXXXXX")#(otherwise we would be dealing with the square root of a negative value)

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