How do you find the domain and range of #f(x) = - sqrt2 / (x² - 16)#?

1 Answer
Jul 11, 2017

Answer:

Domain is all value of #x# other than #-4# and #4#.

Range is all values except those between #0# and #sqrt2/16# i.e. #(0,sqrt2/16)#

Explanation:

AS #f(x)=-sqrt2/(x^2-16)=-sqrt2/((x+4)(x-4))#,

it is quite apparent that #x# cannot take values #-4# and #4#.

Hence domain of #x# is all values other than #-4# and #4#.

Further, when #x# lies between #-4# and #4# i.e. #|x|<4#,

the denominator is always negative i.e. #f(x)# is positive and maximum value of denominator is when #x=0#, and then #f(x)=sqrt2/16#. Hence this is minimum value of #f(x)# in this range.

When #|x|>4#, #f(x)# is always negative and when #x->oo#,

#f(x)=-sqrt2/(x^2-16)=-(sqrt2/x^2)/(1-16/x^2)# i.e. #f(x)->0#

Hence range of #f(x)# is all values except those between #0# and #sqrt2/16#

graph{-sqrt2/(x^2-16) [-5, 5, -2.5, 2.5]}