determine if the given function has any points of discontinuity. explain your reasoning. f(x) = (x² - b²)/(x…

determine if the given function has any points of discontinuity. explain your reasoning. f(x) = (x² - b²)/(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.

determine if the given function has any points of discontinuity. explain your reasoning. f(x) = (x² - b²)/(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.

Answer

Explanation:

Step1: Recall continuity condition

A function $y = f(x)$ is discontinuous where the denominator is 0. For $f(x)=\frac{x^{2}-b^{2}}{x - b}$, set the denominator equal to 0. $x - b=0$

Step2: Solve for x

Solving $x - b = 0$ gives $x = b$. Also, factor the numerator $x^{2}-b^{2}=(x + b)(x - b)$. So $f(x)=\frac{(x + b)(x - b)}{x - b}$, and for $x\neq b$, $f(x)=x + b$. But at $x = b$, the original - form of the function is undefined due to division by zero.

Answer:

There is a point of discontinuity at $x = b$ because the denominator has the factor $x - b$.