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

1 Answer
Jul 11, 2017

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]}