What is the domain and range of #y = y = (x^2 - 1) / (x+1)#?

Question

Kennedy Adams · Accepted Answer

a) #y=(x^2-1)/(x+1)= (x-1)(x+1)/(x+1)=x-1#
b) Domain: #ℝ = {x | -∞ < x <∞} # All Real x are possible
c) Range: #ℝ = {f(x)=y | -∞ < f(x) <∞}# All Real y are possible

Given: #y=(x^2-1)/(x+1)# Required the Domain and range: Solution Strategy: a) Simplify the function, #y=f(x)# b) Domain: identify all possible value of #x# c) Range: Identify all possible results of the function, #f(x)# a) #y=(x^2-1)/(x+1)= (x-1)(x+1)/(x+1)=x-1# b) Domain: #ℝ = {x | -∞ < x <∞} # All Real x are possible c) Range: #ℝ = {f(x)=y | -∞ < f(x) <∞}# All Real y are possible

Nathan Axel · Answer

undefined The domain of the function $ y = \frac{x^2 - 1}{x+1} $ is all real numbers except for $ x = -1 $, because division by zero is undefined. The range of the function includes all real numbers except for $ y = -1 $, as the function approaches negative infinity as $ x $ approaches positive infinity and approaches positive infinity as $ x $ approaches negative infinity.

Julia Adams · Answer

undefined The domain of the function $ y = \frac{x^2 - 1}{x + 1} $ is all real numbers except $ x = -1 $ because the denominator cannot be zero.

To find the range, analyze the behavior of the function as $ x $ approaches positive infinity and negative infinity. As $ x $ approaches positive infinity, $ y $ approaches positive infinity, and as $ x $ approaches negative infinity, $ y $ approaches negative infinity. Therefore, the range of the function is all real numbers except $ y = -1 $.