What is the domain and range of #m(x) = 5/(x^2+9)#??

1 Answer
Jul 24, 2018

Domain - all reals #x#
Range - #0< y <= 5/9#

Explanation:

Looking at the graph, you can immediately tell that it is an even function and and the graph is positive

Even function is when #f(x)=f(-x)#
#f(x)=5/(x^2+9)#
#f(-x)=5/((-x)^2+9)=5/(x^2+9)=f(x)#
Therefore, #5/(x^2+9)# is an even function

It is positive because #x^2+9# is always positive for all real integers

We also know that there is no x-intercept but there is a y-intercept at #(0,5/9)#

Drawing the graph, we can see that:
Domain - all reals #x#
Range - #0< y <= 5/9#

For the range, #y !=0# because the graph is approaching the asymptote #y=0# so it will never touch the line #y=0#

The graph is below
graph{5/(x^2+9) [-10, 10, -5, 5]}