Question

determine if the given function has any points of discontinuity. explain your reasoning.

p(x)=\frac{x^{2}-b^{2}}{x - b}

there is a point of discontinuity at ( x = b ) because the denominator has the factor ( x - b ).

there are points of discontinuity at both ( x = - b ) and ( x = b ) because the numerator has factors of ( x + b ) and ( x - b ).

there is a point of discontinuity at ( x = - b ) only because the factor of ( x - b ) is common to both the numerator and denominator.

there is a point of discontinuity at ( x = b ) only because the factor of ( x - b ) is common to both the numerator and denominator and factors out.

Explanation:

Step1: Factor the numerator

Use the difference - of - squares formula \(a^{2}-b^{2}=(a + b)(a - b)\). For \(P(x)=\frac{x^{2}-b^{2}}{x - b}\), the numerator \(x^{2}-b^{2}=(x + b)(x - b)\). So \(P(x)=\frac{(x + b)(x - b)}{x - b}\).

Step2: Simplify the function (with a domain restriction)

Cancel out the common factor \((x - b)\) (for \(x
eq b\)). The simplified function is \(y=x + b\) with the domain \(x
eq b\).

Step3: Analyze the discontinuity

A rational function \(y=\frac{f(x)}{g(x)}\) has a discontinuity when \(g(x)=0\). Here, \(g(x)=x - b\). When we cancel the common factor \((x - b)\) (a removable discontinuity), the only value that makes the original function undefined (before simplification) is \(x = b\). The value \(x=-b\) does not make the original function undefined in a non - removable way. If we substitute \(x=-b\) into the simplified function \(y=x + b\) (after accounting for the domain restriction \(x
eq b\)), there is no issue related to the original function's denominator.

Answer:

There is a point of discontinuity at \(x = b\) only because the factor of \(x - b\) is common to both the numerator and denominator and factors out.