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

determine if the given function has any points of discontinuity. explain your reasoning.\n\n p(x)=\frac{x^{2}-b^{2}}{x - b} \n\nthere is a point of discontinuity at ( x = b ) because the denominator has the factor ( x - b ).\n\nthere are points of discontinuity at both ( x = - b ) and ( x = b ) because the numerator has factors of ( x + b ) and ( x - b ).\n\nthere is a point of discontinuity at ( x = - b ) only because the factor of ( x - b ) is common to both the numerator and denominator.\n\nthere 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.

determine if the given function has any points of discontinuity. explain your reasoning.\n\n p(x)=\frac{x^{2}-b^{2}}{x - b} \n\nthere is a point of discontinuity at ( x = b ) because the denominator has the factor ( x - b ).\n\nthere are points of discontinuity at both ( x = - b ) and ( x = b ) because the numerator has factors of ( x + b ) and ( x - b ).\n\nthere is a point of discontinuity at ( x = - b ) only because the factor of ( x - b ) is common to both the numerator and denominator.\n\nthere 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.

Answer

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.

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\neq b)). The simplified function is (y=x + b) with the domain (x\neq 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\neq b)), there is no issue related to the original function's denominator.