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Find

a) any critical values and
b) any relative extrema.

[tex]\[ f(x) = x^2 + 2x + 1 \][/tex]

a) Select the correct choice below and, if necessary, fill in the answer box within your choice.

A. The critical value(s) of the function is/are [tex]\(\boxed{-1}\)[/tex].
(Use a comma to separate answers as needed.)

B. The function has no critical values.

b) Select the correct choice below and, if necessary, fill in the answer box(es) within your choice.

A. The relative minimum point(s) is/are [tex]\(\boxed{(-1, 0)}\)[/tex] and there are no relative maximum points.
(Simplify your answer. Type an ordered pair, using integers or fractions. Use a comma to separate answers.)

B. The relative minimum point(s) is/are [tex]\(\boxed{(-1, 0)}\)[/tex] and the relative maximum point(s) is/are [tex]\(\boxed{-1}\)[/tex].
(Simplify your answers. Type ordered pairs, using integers or fractions. Use a comma to separate answers.)

C. The relative maximum point(s) is/are [tex]\(\boxed{\square}\)[/tex] and there are no relative minimum points.

Sagot :

GinnyAnsweridn
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.
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