Sure, let's analyze the given function [tex]\( f(x) = x^2 + 2x + 1 \)[/tex] step-by-step.
### Critical Values
1. First, we need to find the first derivative of the function [tex]\( f(x) \)[/tex]. The function is [tex]\( f(x) = x^2 + 2x + 1 \)[/tex].
Applying the power rule:
[tex]\[
f'(x) = \frac{d}{dx}(x^2 + 2x + 1) = 2x + 2
\][/tex]
2. Set the first derivative equal to zero to find critical points:
[tex]\[
2x + 2 = 0
\][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[
2x = -2 \implies x = -1
\][/tex]
Therefore, the critical value is [tex]\( x = -1 \)[/tex].
### Relative Extrema
1. Next, we need to determine the nature of the critical value using the second derivative test.
We find the second derivative of [tex]\( f(x) \)[/tex]:
[tex]\[
f''(x) = \frac{d}{dx}(2x + 2) = 2
\][/tex]
2. Evaluate the second derivative at the critical value:
[tex]\[
f''(-1) = 2
\][/tex]
Since [tex]\( f''(-1) = 2 \)[/tex] which is greater than zero, it indicates that [tex]\( f(x) \)[/tex] is concave up at [tex]\( x = -1 \)[/tex].
Therefore, the critical value [tex]\( x = -1 \)[/tex] corresponds to a relative minimum.
### Conclusion
- Critical Value: The critical value of [tex]\( f(x) \)[/tex] is [tex]\( x = -1 \)[/tex].
- Relative Extrema: The relative extremum of [tex]\( f(x) \)[/tex] at [tex]\( x = -1 \)[/tex] is a relative minimum.
