How do you find the domain and range of #(x^2-64)/(x-8)#?

1 Answer
Apr 1, 2016

Answer:

Domain: #RR "\" { 8 } = (-oo, 8) uu (8, oo)#

Range: #RR "\" { 16 } = (-oo, 16) uu (16, oo)#

Explanation:

#f(x) = (x^2-64)/(x-8) = ((x-8)(x+8))/(x-8) = x+8#

with exclusion #x != 8#

When #x=8#, both the numerator and denominator are #0# resulting in an undefined value. So #x=8# is not in the domain.

#f(x)# is well defined for all other values of #x#

So the domain is #RR "\" { 8 } = (-oo, 8) uu (8, oo)#

The range is the set of values that #f(x)# takes for #x in RR "\" { 8 }#, namely #(-oo, 16) uu (16, oo)#.

Note that #f(x)# can take any value except #16#