To solve the function [tex]\( f(x) = \frac{\sqrt{x^2 + 2x + 1}}{x^2 - 2x + 1} \)[/tex], let's simplify both the numerator and the denominator step-by-step.
### Step 1: Simplify the Numerator
First, we look at the expression under the square root in the numerator:
[tex]\[ x^2 + 2x + 1 \][/tex]
This can be factored as:
[tex]\[ x^2 + 2x + 1 = (x + 1)^2 \][/tex]
So, the numerator becomes:
[tex]\[ \sqrt{x^2 + 2x + 1} = \sqrt{(x + 1)^2} \][/tex]
Since the square root of a squared term is the absolute value of the term, we get:
[tex]\[ \sqrt{(x + 1)^2} = |x + 1| \][/tex]
### Step 2: Simplify the Denominator
Next, we examine the denominator:
[tex]\[ x^2 - 2x + 1 \][/tex]
This can also be factored as:
[tex]\[ x^2 - 2x + 1 = (x - 1)^2 \][/tex]
### Step 3: Write the Simplified Function
Using these simplifications, we rewrite the function [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = \frac{|x + 1|}{(x - 1)^2} \][/tex]
### Step 4: Analyze the Function for Different Values of [tex]\( x \)[/tex]:
Now, let's evaluate the function for specific values of [tex]\( x \)[/tex] to better understand its behavior.
1. For [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = \frac{|-2 + 1|}{(-2 - 1)^2} = \frac{|-1|}{(-3)^2} = \frac{1}{9} \][/tex]
2. For [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = \frac{|-1 + 1|}{(-1 - 1)^2} = \frac{0}{(-2)^2} = 0 \][/tex]
3. For [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{|0 + 1|}{(0 - 1)^2} = \frac{1}{(-1)^2} = 1 \][/tex]
4. For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = \frac{|1 + 1|}{(1 - 1)^2} \][/tex]
Here, the denominator becomes 0:
[tex]\[ (1 - 1)^2 = 0 \][/tex]
Since division by 0 is undefined, [tex]\( f(1) \)[/tex] is undefined.
5. For [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = \frac{|2 + 1|}{(2 - 1)^2} = \frac{3}{1} = 3 \][/tex]
### Step 5: Summary of Results
From the evaluations above, the results are:
- [tex]\( f(-2) = \frac{1}{9} \)[/tex]
- [tex]\( f(-1) = 0 \)[/tex]
- [tex]\( f(0) = 1 \)[/tex]
- [tex]\( f(1) \)[/tex] is undefined
- [tex]\( f(2) = 3 \)[/tex]
These calculations confirm the simplified form of the function. For any general [tex]\( x \)[/tex],
[tex]\[ f(x) = \frac{|x + 1|}{(x - 1)^2} \][/tex]
provides the correct evaluation given the stipulations of the numerator and denominator.
