How do you find the domain and range of #f(x)=(x^2+2x)/(x+1) #?

Question

Ethan Adkins · Accepted Answer

The domain of #f(x)# is #d_f(x)=RR-{-1}#
The range is #f(x) in RR# Our role is #f(x)=(x^2+2x)/(x+1)# As we cannot divide by #0#, #x+1!=0# Consequently, The domain of #f(x)# is #D_f(x)=RR-{-1}# Let, #y=(x^2+2x)/(x+1)# So, #y(x+1)=x^2+2x# #x^2+2x-xy-y=0# #x^2+(2-y)x-y=0# For the solutions to this quadratic equation to exist, the discriminant #Delta>=0#, #=>#, #b^2-4ac>=0# #(2-y)^2-4*1+(-y)>=0# #4-4y+y^2+4y>=0# #y^2+4>=0# #AA y in RR#, #y^2+4>=0# The range is #y in RR# graph{(x^2+2x)/(x+1) [-12.66, 12.65, -6.33, 6.33]}

Nathan Axel · Answer

undefined To find the domain of f(x), we need to identify any values of x that would result in division by zero. The only value of x that would make the denominator zero is when x = -1. Therefore, the domain of f(x) is all real numbers except x = -1.

To find the range, we can analyze the behavior of the function as x approaches positive infinity and negative infinity. As x approaches positive or negative infinity, the function approaches the value of 1. Therefore, the range of f(x) is all real numbers except f(x) ≠ 1.