Question

find the domain and range of f(x) = 1/(x^2-1) please show me the steps to find the range.

Answer 1

One way to find the range is to find the inverse function. The *domain* of the inverse is the *range* of the original. Domain is easier to find than range.

Answer 2

\[\Large y = \frac{1} {x^2-1}\] \[\Large x^2-1 = \frac{1} {y}\] \[\Large x =\pm \sqrt { \frac{1} {y}+1}\]
It will be where \[\large \frac{ 1 }{ y} +1 \ge 0\]

Answer 3

i want to understand how you write 1/y+1 >= 0 in the interval notation

Answer 4

i can't find an easy way to show it algebraically...\[\large \frac{ 1 }{ y} \ge -1\]but it'll be

when y>0 (ie when y is positive 1/y is always >-1)

when y=<-1 (remember if y is something like -0.5, 1/-0.5 is -2. But if y is -2, 1/-2 is > -1)

Answer 5

graph confirms it https://www.google.com/search?q=1%2F(x%5E2-1)&oq=1%2F(x%5E2-1)&aqs=chrome..69i57j69i58&sourceid=chrome&espv=210&es_sm=93&ie=UTF-8