determine if the given function has any points of discontinuity. explain your reasoning.\n$f(x)=\frac{x^{2}-b…

determine if the given function has any points of discontinuity. explain your reasoning.\n$f(x)=\frac{x^{2}-b^{2}}{x - b}$\no there is a point of discontinuity at $x = b$ because the denominator has the factor $x - b$.\no there are points of discontinuity at both $x=-b$ and $x = b$ because the numerator has factors of $x + b$ and $x - b$.\no there is a point of discontinuity at $x=-b$ only because the factor of $x - b$ is common to both the numerator and denominator.\no 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.\n$f(x)=\frac{x^{2}-b^{2}}{x - b}$\no there is a point of discontinuity at $x = b$ because the denominator has the factor $x - b$.\no there are points of discontinuity at both $x=-b$ and $x = b$ because the numerator has factors of $x + b$ and $x - b$.\no there is a point of discontinuity at $x=-b$ only because the factor of $x - b$ is common to both the numerator and denominator.\no 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: Factor the numerator

We know that $x^{2}-b^{2}=(x + b)(x - b)$ by the difference - of - squares formula. So $f(x)=\frac{(x + b)(x - b)}{x - b}$.

Step2: Simplify the function

For $x\neq b$, we can cancel out the common factor $(x - b)$ in the numerator and denominator, and we get $f(x)=x + b$ for $x\neq b$.

Step3: Analyze the discontinuity

The original function $f(x)=\frac{x^{2}-b^{2}}{x - b}$ is undefined when the denominator is zero. Setting $x - b=0$, we find $x = b$. Since we can cancel the factor $(x - b)$ and the function has a removable discontinuity at $x = b$.

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.