To find the critical values and relative extrema of the given function [tex]\( f(x) = x^2 + 2x + 1 \)[/tex], follow these steps:
### Step 1: Find the First Derivative
To find the critical values, we first need to find the first derivative of the function. The given function is:
[tex]\[ f(x) = x^2 + 2x + 1 \][/tex]
Taking the derivative with respect to [tex]\( x \)[/tex], we get:
[tex]\[ f'(x) = \frac{d}{dx}(x^2 + 2x + 1) = 2x + 2 \][/tex]
### Step 2: Find Critical Values
Critical values are found by setting the first derivative equal to zero and solving for [tex]\( x \)[/tex]:
[tex]\[ 2x + 2 = 0 \][/tex]
[tex]\[ 2x = -2 \][/tex]
[tex]\[ x = -1 \][/tex]
Thus, the critical value is [tex]\( x = -1 \)[/tex].
So, the correct choice for part (a) is:
A. The critical value(s) of the function is/are -1.
### Step 3: Determine Relative Extrema
Next, we need to determine whether the critical point is a relative minimum, maximum, or neither. To do this, we use the second derivative test.
First, find the second derivative of the function:
[tex]\[ f''(x) = \frac{d}{dx}(2x + 2) = 2 \][/tex]
Evaluate the second derivative at the critical point [tex]\( x = -1 \)[/tex]:
[tex]\[ f''(-1) = 2 \][/tex]
Since [tex]\( f''(-1) = 2 > 0 \)[/tex], the second derivative is positive, indicating that the function has a relative minimum at [tex]\( x = -1 \)[/tex].
To find the value of the function at this critical point:
[tex]\[ f(-1) = (-1)^2 + 2(-1) + 1 = 1 - 2 + 1 = 0 \][/tex]
Thus, the relative minimum point is [tex]\((-1, 0)\)[/tex], and there are no relative maximum points.
So, the correct choice for part (b) is:
A. The relative minimum point(s) is/are (-1, 0) and there are no relative maximum points.
