Question
Use algebraic and/or graphical methods to solve the inequality.
(xminus14)(xminus17)(xplus14)greater than or equals0
Question content area bottom
Part 1
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A.
The solution(s) is/are xequals
enter your response here.
(Use a comma to separate answers as needed.)
B.
The solutions are the values of x in the interval
enter your response here.
(Type your answer in interval notation.)
C.
There is no solution.
(xminus14)(xminus17)(xplus14)greater than or equals0
Question content area bottom
Part 1
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A.
The solution(s) is/are xequals
enter your response here.
(Use a comma to separate answers as needed.)
B.
The solutions are the values of x in the interval
enter your response here.
(Type your answer in interval notation.)
C.
There is no solution.
Answers
GPT-4o mini
To solve the inequality \((x - 14)(x - 17)(x + 14) \geq 0\), we first find the critical points by setting each factor equal to zero.
1. \(x - 14 = 0 \) gives \(x = 14\)
2. \(x - 17 = 0\) gives \(x = 17\)
3. \(x + 14 = 0\) gives \(x = -14\)
Thus, the critical points are \(x = -14\), \(x = 14\), and \(x = 17\).
Next, we'll determine the sign of the expression \((x - 14)(x - 17)(x + 14)\) in the intervals defined by these critical points:
- \((- \infty, -14)\)
- \((-14, 14)\)
- \((14, 17)\)
- \((17, \infty)\)
### Test the intervals:
1. **Interval \((- \infty, -14)\)**: Choose \(x = -15\).
\[
(-15 - 14)(-15 - 17)(-15 + 14) = (-29)(-32)(-1) < 0
\]
2. **Interval \((-14, 14)\)**: Choose \(x = 0\).
\[
(0 - 14)(0 - 17)(0 + 14) = (-14)(-17)(14) > 0
\]
3. **Interval \((14, 17)\)**: Choose \(x = 15\).
\[
(15 - 14)(15 - 17)(15 + 14) = (1)(-2)(29) < 0
\]
4. **Interval \((17, \infty)\)**: Choose \(x = 18\).
\[
(18 - 14)(18 - 17)(18 + 14) = (4)(1)(32) > 0
\]
### Summary of intervals:
- In \((- \infty, -14)\), the expression is negative.
- In \((-14, 14)\), the expression is positive.
- In \((14, 17)\), the expression is negative.
- In \((17, \infty)\), the expression is positive.
### Include the critical points:
At the critical points \(x = -14\), \(x = 14\), and \(x = 17\), the expression equals zero:
- \((x - 14)(x - 17)(x + 14) = 0\)
### Solution:
From our tests, the expression is non-negative in the intervals:
- \([-14, 14]\)
- \([17, \infty)\)
Combined, the solution in interval notation is:
\[
[-14, 14] \cup [17, \infty)
\]
### Final answer:
B. The solutions are the values of \(x\) in the interval \([-14, 14] \cup [17, \infty)\).
1. \(x - 14 = 0 \) gives \(x = 14\)
2. \(x - 17 = 0\) gives \(x = 17\)
3. \(x + 14 = 0\) gives \(x = -14\)
Thus, the critical points are \(x = -14\), \(x = 14\), and \(x = 17\).
Next, we'll determine the sign of the expression \((x - 14)(x - 17)(x + 14)\) in the intervals defined by these critical points:
- \((- \infty, -14)\)
- \((-14, 14)\)
- \((14, 17)\)
- \((17, \infty)\)
### Test the intervals:
1. **Interval \((- \infty, -14)\)**: Choose \(x = -15\).
\[
(-15 - 14)(-15 - 17)(-15 + 14) = (-29)(-32)(-1) < 0
\]
2. **Interval \((-14, 14)\)**: Choose \(x = 0\).
\[
(0 - 14)(0 - 17)(0 + 14) = (-14)(-17)(14) > 0
\]
3. **Interval \((14, 17)\)**: Choose \(x = 15\).
\[
(15 - 14)(15 - 17)(15 + 14) = (1)(-2)(29) < 0
\]
4. **Interval \((17, \infty)\)**: Choose \(x = 18\).
\[
(18 - 14)(18 - 17)(18 + 14) = (4)(1)(32) > 0
\]
### Summary of intervals:
- In \((- \infty, -14)\), the expression is negative.
- In \((-14, 14)\), the expression is positive.
- In \((14, 17)\), the expression is negative.
- In \((17, \infty)\), the expression is positive.
### Include the critical points:
At the critical points \(x = -14\), \(x = 14\), and \(x = 17\), the expression equals zero:
- \((x - 14)(x - 17)(x + 14) = 0\)
### Solution:
From our tests, the expression is non-negative in the intervals:
- \([-14, 14]\)
- \([17, \infty)\)
Combined, the solution in interval notation is:
\[
[-14, 14] \cup [17, \infty)
\]
### Final answer:
B. The solutions are the values of \(x\) in the interval \([-14, 14] \cup [17, \infty)\).