How do you find the range of #f(x)= 1/(x^2 + 9)#? Algebra Expressions, Equations, and Functions Domain and Range of a Function 1 Answer bp Sep 16, 2015 Range #{ y in RR#, 0< y<= 1/9}# Explanation: Solve # y= 1/(x^2+9)# for x #x^2= (1-9y)/y#, #x= sqrt((1-9y)/y)# This implies that #y !=0# and #1-9y>=0#, that is #y <= 1/9# Hence range of f(x) would be #{ y in RR,0< y <= 1/9}# Answer link Related questions How do you determine if (-1, 4), (2, 8), (-1, 5) is a function? What is the domain for #f(x)=2x-4#? What is the domain and range for (3,1), (1,-4), and (2, 8)? What is the domain and range of a linear function? Is domain the independent or dependent variable? How do you find the domain and range of a function in interval notation? How do you find domain and range of a rational function? How do you find domain and range of a quadratic function? How do you determine the domain and range of a function? What is Domain and Range of a Function? See all questions in Domain and Range of a Function Impact of this question 5178 views around the world You can reuse this answer Creative Commons License