Answer :
Let's analyze the given function step-by-step:
[tex]\[ f(x) = \frac{x^2 - b^2}{x - b} \][/tex]
### Step 1: Factoring the Numerator
Notice that the numerator [tex]\( x^2 - b^2 \)[/tex] can be factored using the difference of squares formula:
[tex]\[ x^2 - b^2 = (x - b)(x + b) \][/tex]
So, we rewrite the function as:
[tex]\[ f(x) = \frac{(x - b)(x + b)}{x - b} \][/tex]
### Step 2: Simplification
For [tex]\( x \neq b \)[/tex], we can cancel out the common factor [tex]\((x - b)\)[/tex] from the numerator and the denominator:
[tex]\[ f(x) = x + b \quad \text{for} \quad x \neq b \][/tex]
### Step 3: Identifying Points of Discontinuity
The original form [tex]\(\frac{x^2 - b^2}{x - b}\)[/tex] has a denominator that becomes zero when [tex]\( x = b \)[/tex]. Thus, the function is undefined at [tex]\( x = b \)[/tex]. This means there is a discontinuity at [tex]\( x = b \)[/tex].
### Step 4: Analyzing Solution
The simplified function [tex]\( f(x) = x + b \)[/tex] is valid for all other values of [tex]\( x \)[/tex], except where [tex]\( x = b \)[/tex].
### Conclusion
Therefore, the function [tex]\( f(x) \)[/tex] has a point of discontinuity at [tex]\( x = b \)[/tex] because at this point the denominator [tex]\( x - b \)[/tex] is zero, causing the original function to be undefined.
There is no discontinuity at [tex]\( x = -b \)[/tex] because this point does not cause the denominator to be zero.
The correct reasoning for the discontinuity is:
There is a point of discontinuity at [tex]\( x = b \)[/tex] only because the factor of [tex]\( x - b \)[/tex] is common to both the numerator and denominator and factors out, leaving the function undefined at [tex]\( x = b \)[/tex].
[tex]\[ f(x) = \frac{x^2 - b^2}{x - b} \][/tex]
### Step 1: Factoring the Numerator
Notice that the numerator [tex]\( x^2 - b^2 \)[/tex] can be factored using the difference of squares formula:
[tex]\[ x^2 - b^2 = (x - b)(x + b) \][/tex]
So, we rewrite the function as:
[tex]\[ f(x) = \frac{(x - b)(x + b)}{x - b} \][/tex]
### Step 2: Simplification
For [tex]\( x \neq b \)[/tex], we can cancel out the common factor [tex]\((x - b)\)[/tex] from the numerator and the denominator:
[tex]\[ f(x) = x + b \quad \text{for} \quad x \neq b \][/tex]
### Step 3: Identifying Points of Discontinuity
The original form [tex]\(\frac{x^2 - b^2}{x - b}\)[/tex] has a denominator that becomes zero when [tex]\( x = b \)[/tex]. Thus, the function is undefined at [tex]\( x = b \)[/tex]. This means there is a discontinuity at [tex]\( x = b \)[/tex].
### Step 4: Analyzing Solution
The simplified function [tex]\( f(x) = x + b \)[/tex] is valid for all other values of [tex]\( x \)[/tex], except where [tex]\( x = b \)[/tex].
### Conclusion
Therefore, the function [tex]\( f(x) \)[/tex] has a point of discontinuity at [tex]\( x = b \)[/tex] because at this point the denominator [tex]\( x - b \)[/tex] is zero, causing the original function to be undefined.
There is no discontinuity at [tex]\( x = -b \)[/tex] because this point does not cause the denominator to be zero.
The correct reasoning for the discontinuity is:
There is a point of discontinuity at [tex]\( x = b \)[/tex] only because the factor of [tex]\( x - b \)[/tex] is common to both the numerator and denominator and factors out, leaving the function undefined at [tex]\( x = b \)[/tex].
