Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
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].
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.