What is the domain and range of #f(x)= x^2 - 6x + 8#?

2 Answers
Aug 9, 2015

Answer:

Domain: #x in R# or #{x: -oo<=x<=oo}#. #x# can take up any real values.

Range: #{f(x):-1<=f(x)<=oo}#

Explanation:

Domain:

#f(x)# is a quadratic equation and any values of #x# will give a real value of #f(x)#.

The function does not converge to a certain value ie: #f(x)=0# when #x->oo#

Your domain is #{x: -oo<=x<=oo}#.

Range:

Method 1-
Use completing the square method:
#x^2-6x+8=(x-3)^2-1#
Hence you minimum point is #(3,-1)#. It is a minimum point because the graph is a "u" shape (coefficient of #x^2# is positive).

Method 2-
Differentiate :
#(df(x))/(dx)=2x-6#.

Let#(df(x))/(dx)=0#

Therefore, #x=3# and #f(3)=-1#
Minimum point is #(3,-1)#.
It is a minimum point because the graph is a "u" shape (coefficient of #x^2# is positive).

Your range takes values between #-1 and oo#

Aug 9, 2015

Answer:

Domain #(-oo,+oo)#
Range #[-1, +oo)#

Explanation:

It is a polynomial function, its domain is all real numbers. In interval notation this can be expressed as #(-oo, +oo)#
For finding its range, we can solve the equation y= #x^2-6x+8# for x first as follows:

#y= (x-3)^2 -1#,
#(x-3)^2 = y+1#
x-3= #+-sqrt(y+1)#
x= 3#+-sqrt(y+1)#. It is obvious from this that y#>=-1#

Hence range is #y>=-1#. In interval notation this can be expressed as# [-1, +oo)#