Find the x-values of all points where the function has any relative extrema. Find the value(s) of any relative extrema. f(x)=x²- -x+3 Select the correct choice below and, if necessary, fill in any answer boxes within your choice. at x = OA. There are no relative maxima. The function has a relative minimum of (Use a comma to separate answers as needed.) at x = OB. There are no relative minima. The function has a relative maximum of (Use a comma to separate answers as needed.) OC. The function has a relative maximum of (Use a comma to separate answers as needed.) OD. There are no relative extrema. at x = and a relative minimum of at x =

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
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Chapter1: Expressions And Functions
Section: Chapter Questions
Problem 74SGR
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### Finding Relative Extrema

This problem requires determining the x-values of all points where the function \( f(x) = x^2 - x + 3 \) has any relative extrema, as well as finding the values of any relative extrema.

**Problem Statement:**

Find the x-values of all points where the function has any relative extrema. Find the value(s) of any relative extrema.
\[ f(x) = x^2 - x + 3 \]

**Question Options:**

1. **Option A:**
   - There are no relative maxima.
   - The function has a relative minimum of [ ] at \( x = [ ] \).
   - (Use a comma to separate answers as needed.)

2. **Option B:**
   - There are no relative minima.
   - The function has a relative maximum of [ ] at \( x = [ ] \).
   - (Use a comma to separate answers as needed.)

3. **Option C:**
   - The function has a relative maximum of [ ] at \( x = [ ] \) and a relative minimum of [ ] at \( x = [ ] \).
   - (Use a comma to separate answers as needed.)

4. **Option D:**
   - There are no relative extrema.

**Instructions:**

Select the correct choice below and, if necessary, fill in any answer boxes within your choice.

This problem is typically solved by finding the first derivative of the function, setting it equal to zero to find critical points, and then using the second derivative test to classify these points as relative maxima, minima, or points of inflection.

### Steps to Solve:

1. **Find the First Derivative of \( f(x) \)**:
   \[ f'(x) = 2x - 1 \]

2. **Set the First Derivative Equal to Zero**:
   \[ 2x - 1 = 0 \]
   \[ x = \frac{1}{2} \]

3. **Find the Second Derivative of \( f(x) \)**:
   \[ f''(x) = 2 \]

4. **Classify the Critical Point Using the Second Derivative**:
   Since \( f''(x) = 2 > 0 \), the function is concave up at \( x = \frac{1}{2} \), indicating a relative minimum.

5. **Calculate the Relative Minimum**
Transcribed Image Text:### Finding Relative Extrema This problem requires determining the x-values of all points where the function \( f(x) = x^2 - x + 3 \) has any relative extrema, as well as finding the values of any relative extrema. **Problem Statement:** Find the x-values of all points where the function has any relative extrema. Find the value(s) of any relative extrema. \[ f(x) = x^2 - x + 3 \] **Question Options:** 1. **Option A:** - There are no relative maxima. - The function has a relative minimum of [ ] at \( x = [ ] \). - (Use a comma to separate answers as needed.) 2. **Option B:** - There are no relative minima. - The function has a relative maximum of [ ] at \( x = [ ] \). - (Use a comma to separate answers as needed.) 3. **Option C:** - The function has a relative maximum of [ ] at \( x = [ ] \) and a relative minimum of [ ] at \( x = [ ] \). - (Use a comma to separate answers as needed.) 4. **Option D:** - There are no relative extrema. **Instructions:** Select the correct choice below and, if necessary, fill in any answer boxes within your choice. This problem is typically solved by finding the first derivative of the function, setting it equal to zero to find critical points, and then using the second derivative test to classify these points as relative maxima, minima, or points of inflection. ### Steps to Solve: 1. **Find the First Derivative of \( f(x) \)**: \[ f'(x) = 2x - 1 \] 2. **Set the First Derivative Equal to Zero**: \[ 2x - 1 = 0 \] \[ x = \frac{1}{2} \] 3. **Find the Second Derivative of \( f(x) \)**: \[ f''(x) = 2 \] 4. **Classify the Critical Point Using the Second Derivative**: Since \( f''(x) = 2 > 0 \), the function is concave up at \( x = \frac{1}{2} \), indicating a relative minimum. 5. **Calculate the Relative Minimum**
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