give the domain and range

Question

give the domain and the range of f(x) = 1/x^2-8

Answer 1

For domain:

 

The only special points are those when denominator is zero. Then the function is not defined and those points shall be excluded from domain.

 

x²-8=0;

x=±2√2;

 

Thus domain, D, of a function is D:x∈(-∞;-2√2)∪(-2√2;2√2)∪(2√2;∞).

 

For the range:

 

Consider the denominator. Since x² is always non-negative, x²-8 is always greater or equal to -8. When the x approaches -2√2 from the right of 2√2 from the left, denominator is negative and goes to zero, therefore the whole function goes to minus infinity. So on the interval (-2√2;2√2) the function f(x)=1/(x²-8) goes from -1/8 to -∞. On two other intervals the function goes to +infinity  when x approaches -2√2 from the left or 2√2 from the right. When x goes to ±∞, denominator goes to +∞ and the whole function, f(x), goes to zero, but never attains that value. Thus, the range of this function is:

 

R: f(x)∈(-∞;-1/8]∪(0;∞).