Find the range of the function \( f(x) = \frac{x^2 + 1}{x^2 - 1} \) as \( x \) ranges over all real numbers such that \( x \neq \pm 1 \).
["# Finding the Range of the Function ( f(x) = \frac{x^2 + 1}{x^2 - 1} ) for ( x <br/>\neq \pm 1 )", "Understanding the range of a rational function like ( f(x) = \frac{x^2 + 1}{x^2 - 1} ) helps reveal all possible output values based on valid input values. In this article, we’ll find and explain the full range of ( f(x) ) for all real ( x ) except ( x = \pm 1 ), where the function is undefined.", "---", "## The Function Definition", "Consider the function:", "[\nf(x) = \frac{x^2 + 1}{x^2 - 1}, \quad x <br/>\neq \pm 1\n]", "Since both numerator and denominator are polynomials in ( x^2 ), and the denominator is zero only at ( x = \pm 1 ), the domain excludes only ( x = 1 ) and ( x = -1 ).", "---", "## Goal", "Find all real values ( y ) such that the equation ( y = \frac{x^2 + 1}{x^2 - 1} ) has at least one real solution ( x ) with ( x^2 <br/>\ne 1 ).", "---", "## Step 1: Set Up the Equation and Substitute", "Let:", "[\ny = \frac{x^2 + 1}{x^2 - 1}\n]", "Multiply both sides by ( x^2 - 1 ) (valid since ( x^2 <br/>\ne 1 )):", "[\ny(x^2 - 1) = x^2 + 1\n]", "Expand:", "[\ny x^2 - y = x^2 + 1\n]", "Bring all terms to one side:", "[\ny x^2 - x^2 - y - 1 = 0\n]", "Factor ( x^2 ):", "[\nx^2(y - 1) = y + 1\n]", "Now solve for ( x^2 ):", "[\nx^2 = \frac{y + 1}{y - 1}, \quad \ ext{provided } y <br/>\ne 1\n]", "---", "## Step 2: Analyze Valid Outputs", "Since ( x^2 \geq 0 ) for real ( x ), we require:", "[\n\frac{y + 1}{y - 1} \geq 0\n]", "This inequality determines the possible values of ( y ). We analyze the sign of the rational expression.", "### Critical Points", "The expression changes sign at numerator zero and denominator zero:", "- ( y + 1 = 0 \Rightarrow y = -1 )\n- ( y - 1 = 0 \Rightarrow y = 1 ) (excluded in denominator)", "These divide the real line into intervals:", "1. ( y < -1 )\n2. ( -1 < y < 1 )\n3. ( y > 1 )", "Test each interval:", "- For ( y < -1 ): both numerator and denominator negative → positive → satisfies inequality.\n- For ( -1 < y < 1 ): numerator positive, denominator negative → negative → not allowed.\n- For ( y > 1 ): both numerator and denominator positive → positive → satisfies inequality.", "Thus,", "[\n\frac{y + 1}{y - 1} \geq 0 \quad \ ext{when} \quad y \leq -1 \quad \ ext{or} \quad y > 1\n]", "---", "## Step 3: Check Boundary Values", "We found ( x^2 = \frac{y+1}{y-1} ), and we require ( x^2 \geq 0 ), which matches the earlier inequality. But we must ensure the solution is valid — specifically, ( x^2 <br/>\ne 1 ) to respect the domain ( x <br/>\ne \pm 1 ).", "Can ( \frac{y+1}{y-1} = 1 )?", "Set:", "[\n\frac{y + 1}{y - 1} = 1 \Rightarrow y + 1 = y - 1 \Rightarrow 1 = -1 \quad \ ext{(no solution)}\n]", "So ( x^2 = 1 ) is never achieved for valid ( y ) in the solution set. Thus, the only restriction comes from the inequality.", "---", "## Step 4: Confirm All Values Are Achievable", "We must confirm that every ( y \leq -1 ) or ( y > 1 ) gives a real ( x <br/>\ne \pm 1 ).", "- When ( y \leq -1 ), ( \frac{y+1}{y-1} \geq 0 ) and ( > 0 ) (equal to 0 only when ( y = -1 )), so ( x^2 \geq 0 ).\n For example, when ( y = -1 ):", "[\n x^2 = \frac{-1 + 1}{-1 - 1} = \frac{0}{-2} = 0 \Rightarrow x = 0\n ]", "And ( x = 0 ) is allowed (( x <br/>\ne \pm 1 )).", "- When ( y > 1 ), ( \frac{y+1}{y-1} \in (0,1) ), so ( x^2 < 1 ) → ( |x| < 1 ), valid.", "Hence, every such ( y ) corresponds to valid real ( x ) (except ( x = \pm 1 ), which are not in the domain anyway).", "---", "## Conclusion", "The range of ( f(x) = \frac{x^2 + 1}{x^2 - 1} ) for ( x \in \mathbb{R} \setminus { \pm 1 } ) is all real numbers ( y ) such that ( y \leq -1 ) or ( y > 1 ).", "So:", "[\n\boxed{(-\infty, -1] \cup (1, \infty)}\n]", "This means the function never outputs values in ( (-1, 1) ), and achieves every value ≥ -1 (except -1 when ( x = 0 ), but -1 is included via ( y = -1 )), as well as all values greater than 1.", "---", "### Key Takeaways:", "- The denominator zero at ( x = \pm 1 ) creates vertical asymptotes, but does not restrict the full range.\n- Solving by equating ( y = f(x) ) and analyzing the resulting equation yields the range through rational inequality.\n- The condition ( x^2 \geq 0 ) restricts ( y ) to ( (-\infty, -1] \cup (1, \infty) ).\n- The function attains all permitted values: it approaches ( -\infty ) near ( x = \pm 1 ) from inside the domain, reaches -1 at ( x = 0 ), and increases toward 1 from below as ( |x| \ o \infty ), but never reaches 1.", "Understanding this range helps in graphing, solving equations, and modeling real-world phenomena constrained to continuous domains excluding singular points."]
