Answer :
To determine the solution set of the quadratic inequality [tex]\( f(x) \geq 0 \)[/tex], let's go through the steps to solve a typical quadratic inequality.
Given a quadratic function [tex]\( f(x) = ax^2 + bx + c \)[/tex], we typically solve such inequalities as follows:
1. Find the roots of the quadratic equation [tex]\( f(x) = 0 \)[/tex].
2. Determine the intervals defined by these roots.
3. Test the sign of [tex]\( f(x) \)[/tex] in each interval.
4. Combine the intervals where [tex]\( f(x) \geq 0 \)[/tex].
### Example Quadratic Function
Let's solve the quadratic inequality [tex]\( f(x) = x^2 - 4x + 3 \geq 0 \)[/tex].
1. Find the roots:
We solve the equation [tex]\( x^2 - 4x + 3 = 0 \)[/tex].
This is a standard quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex]. To find the roots, we can either factor the quadratic expression or use the quadratic formula. Let's factor it:
[tex]\[ x^2 - 4x + 3 = (x - 1)(x - 3) = 0 \][/tex]
So, the roots are [tex]\( x = 1 \)[/tex] and [tex]\( x = 3 \)[/tex].
2. Determine the intervals:
The roots divide the real number line into three intervals:
- [tex]\( (-\infty, 1) \)[/tex]
- [tex]\( (1, 3) \)[/tex]
- [tex]\( (3, \infty) \)[/tex]
3. Test the sign in each interval:
We choose a test point in each interval to determine if [tex]\( f(x) \geq 0 \)[/tex] in that interval.
- For the interval [tex]\( (-\infty, 1) \)[/tex], choose [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 0^2 - 4 \cdot 0 + 3 = 3 \quad (\text{positive}) \][/tex]
- For the interval [tex]\( (1, 3) \)[/tex], choose [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 2^2 - 4 \cdot 2 + 3 = 4 - 8 + 3 = -1 \quad (\text{negative}) \][/tex]
- For the interval [tex]\( (3, \infty) \)[/tex], choose [tex]\( x = 4 \)[/tex]:
[tex]\[ f(4) = 4^2 - 4 \cdot 4 + 3 = 16 - 16 + 3 = 3 \quad (\text{positive}) \][/tex]
4. Combine the intervals where [tex]\( f(x) \geq 0 \)[/tex]:
[tex]\( f(x) \geq 0 \)[/tex] in the intervals [tex]\( (-\infty, 1] \)[/tex] and [tex]\( [3, \infty) \)[/tex].
The quadratic function [tex]\( f(x) = x^2 - 4x + 3 \)[/tex] is 0 at the roots [tex]\( x = 1 \)[/tex] and [tex]\( x = 3 \)[/tex], so these points are included in our solution set.
Therefore, the solution set of the quadratic inequality [tex]\( x^2 - 4x + 3 \geq 0 \)[/tex] is:
[tex]\[ (-\infty, 1] \cup [3, \infty) \][/tex]
Given your possible choices:
- [tex]\( \{x \mid x \in \mathbb{R} \} \)[/tex]: This is incorrect, as not all real numbers satisfy the inequality.
- [tex]\( \varnothing \)[/tex]: This is incorrect, as there are real solutions.
- [tex]\( \{x \mid x = -3 \} \)[/tex]: This is incorrect, as it only includes one point which is not even a root of the quadratic.
- [tex]\( \{x \mid x = 3 \} \)[/tex]: This is incorrect, as it only specifies one of the roots.
Since none of the given choices exactly match our derived solution set, it seems like there might be some confusion, or the correct options might not include the precise interval-based solution we've outlined. The accurate representation should be [tex]\( \{x \mid x \leq 1 \text{ or } x \geq 3\} \)[/tex]. Please verify the choices for correctness based on further context or consult your instructor if needed.
Given a quadratic function [tex]\( f(x) = ax^2 + bx + c \)[/tex], we typically solve such inequalities as follows:
1. Find the roots of the quadratic equation [tex]\( f(x) = 0 \)[/tex].
2. Determine the intervals defined by these roots.
3. Test the sign of [tex]\( f(x) \)[/tex] in each interval.
4. Combine the intervals where [tex]\( f(x) \geq 0 \)[/tex].
### Example Quadratic Function
Let's solve the quadratic inequality [tex]\( f(x) = x^2 - 4x + 3 \geq 0 \)[/tex].
1. Find the roots:
We solve the equation [tex]\( x^2 - 4x + 3 = 0 \)[/tex].
This is a standard quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex]. To find the roots, we can either factor the quadratic expression or use the quadratic formula. Let's factor it:
[tex]\[ x^2 - 4x + 3 = (x - 1)(x - 3) = 0 \][/tex]
So, the roots are [tex]\( x = 1 \)[/tex] and [tex]\( x = 3 \)[/tex].
2. Determine the intervals:
The roots divide the real number line into three intervals:
- [tex]\( (-\infty, 1) \)[/tex]
- [tex]\( (1, 3) \)[/tex]
- [tex]\( (3, \infty) \)[/tex]
3. Test the sign in each interval:
We choose a test point in each interval to determine if [tex]\( f(x) \geq 0 \)[/tex] in that interval.
- For the interval [tex]\( (-\infty, 1) \)[/tex], choose [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 0^2 - 4 \cdot 0 + 3 = 3 \quad (\text{positive}) \][/tex]
- For the interval [tex]\( (1, 3) \)[/tex], choose [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 2^2 - 4 \cdot 2 + 3 = 4 - 8 + 3 = -1 \quad (\text{negative}) \][/tex]
- For the interval [tex]\( (3, \infty) \)[/tex], choose [tex]\( x = 4 \)[/tex]:
[tex]\[ f(4) = 4^2 - 4 \cdot 4 + 3 = 16 - 16 + 3 = 3 \quad (\text{positive}) \][/tex]
4. Combine the intervals where [tex]\( f(x) \geq 0 \)[/tex]:
[tex]\( f(x) \geq 0 \)[/tex] in the intervals [tex]\( (-\infty, 1] \)[/tex] and [tex]\( [3, \infty) \)[/tex].
The quadratic function [tex]\( f(x) = x^2 - 4x + 3 \)[/tex] is 0 at the roots [tex]\( x = 1 \)[/tex] and [tex]\( x = 3 \)[/tex], so these points are included in our solution set.
Therefore, the solution set of the quadratic inequality [tex]\( x^2 - 4x + 3 \geq 0 \)[/tex] is:
[tex]\[ (-\infty, 1] \cup [3, \infty) \][/tex]
Given your possible choices:
- [tex]\( \{x \mid x \in \mathbb{R} \} \)[/tex]: This is incorrect, as not all real numbers satisfy the inequality.
- [tex]\( \varnothing \)[/tex]: This is incorrect, as there are real solutions.
- [tex]\( \{x \mid x = -3 \} \)[/tex]: This is incorrect, as it only includes one point which is not even a root of the quadratic.
- [tex]\( \{x \mid x = 3 \} \)[/tex]: This is incorrect, as it only specifies one of the roots.
Since none of the given choices exactly match our derived solution set, it seems like there might be some confusion, or the correct options might not include the precise interval-based solution we've outlined. The accurate representation should be [tex]\( \{x \mid x \leq 1 \text{ or } x \geq 3\} \)[/tex]. Please verify the choices for correctness based on further context or consult your instructor if needed.
