91Ó°ÊÓ

A rational function is a type of function represented by the ratio of two polynomial expressions. It comes in the form \( R(x) = \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are both polynomials, and \( Q(x) eq 0 \). In the given exercise, the rational function is \( f(x) = \frac{x+1}{x^2 - 2x - 3} \). This structure indicates the function involves a division where the numerator is \( x+1 \) and the denominator is \( x^2 - 2x - 3 \). Rational functions are interesting due to their behavior at points where the denominator equals zero. This is the key to finding discontinuities in these functions.

• Numerator: A polynomial that may not significantly affect discontinuity.
• Denominator: The polynomial, whose zeroes are crucial because they can cause the function to be undefined.

Factorization is the process of breaking down a complex algebraic expression into simpler components or factors that, when multiplied together, produce the original expression. In our exercise, the denominator \(x^2 - 2x - 3\) is a quadratic polynomial. To factor it, we look for two numbers that multiply to -3 (the constant term) and add to -2 (the linear coefficient). The correct factor pair is \(-3\) and \(+1\), giving the factorization of \((x-3)(x+1)\).Once factored, our rational function looks like this: \[f(x) = \frac{x+1}{(x-3)(x+1)}\]This makes it easier to analyze and potentially simplify the expression.

Simplification involves reducing an expression to its most basic form. For our function \( f(x) = \frac{x+1}{(x-3)(x+1)} \), simplification can occur by canceling common factors in the numerator and the denominator.Since \( x+1 \) appears in both the numerator and one of the parts of the denominator, they can be canceled out: \[f(x) = \frac{1}{x-3}\]This simplification is only valid when \( x+1 eq 0 \), which means \( x eq -1 \). Thus, while the function has no discontinuity at \( x = -1 \) in the simplified form, \( x = -1 \) is still technically a point of removal discontinuity in the original function.

The zeros of a function are the values of \( x \) that make the function equal to zero. For rational functions, zeros can provide information about where the function might cross the x-axis, but more importantly for this context, knowing the zeros of the denominator helps in identifying where the function is undefined and thus, discontinuous.In our simplified function \( f(x) = \frac{1}{x-3} \), the primary focus for discontinuity is the value that makes the denominator zero. Solving for the zero gives:\[x - 3 = 0 \x = 3\]When \( x = 3 \), the function becomes undefined because division by zero is not possible, resulting in a point of discontinuity at \( x = 3 \). This is an essential aspect when studying rational functions and their continuity properties.