Question

Question #f628a

Answer 1

See below.

Explanation:

`f(x)=2/(x^2-16)`

We can see that the maximum value of the function will occur when the denominator is the minimum positive value, and the minimum value will occur when the denominator is the minimum negative value.

For denominator:

`x^2-16=0=>x=+-4`

Using the limit:

as `x->4^+` , `color(white)(888888)2/(x^2-16)->oo`

as `x->4^-` , `color(white)(888888)2/(x^2-16)->-oo`

as `x->-4^+` , `color(white)(8888)2/(x^2-16)->-oo`

as `x->-4^-` , `color(white)(8888)2/(x^2-16)->oo`

We can see from the above that the range is:

`(-oo , 0)uu(0 , oo)`

or

`{y in RR: y!=0}`

Notice that the function can never equal zero, because the numerator can never be zero.

Graph:

graph{2/(x^2-16) [-10, 10, -5, 5]}