Answer :
To solve the inequality [tex]\( d^2 \geq 2d \)[/tex], let's go through the process step by step:
1. Rewrite the inequality in standard form:
[tex]\[ d^2 - 2d \geq 0 \][/tex]
2. Factor the left-hand side of the inequality:
[tex]\[ d(d - 2) \geq 0 \][/tex]
3. Identify the critical points by setting each factor equal to zero:
[tex]\[ d = 0 \quad \text{and} \quad d = 2 \][/tex]
These critical points divide the number line into three intervals: [tex]\( (-\infty, 0) \)[/tex], [tex]\( (0, 2) \)[/tex], and [tex]\( (2, \infty) \)[/tex].
4. Test a value in each interval to determine where the product is non-negative:
- For the interval [tex]\( (-\infty, 0) \)[/tex], choose [tex]\( d = -1 \)[/tex]:
[tex]\[ (-1)(-1 - 2) = (-1)(-3) = 3 \geq 0 \][/tex]
This interval satisfies the inequality.
- For the interval [tex]\( (0, 2) \)[/tex], choose [tex]\( d = 1 \)[/tex]:
[tex]\[ (1)(1 - 2) = (1)(-1) = -1 < 0 \][/tex]
This interval does not satisfy the inequality.
- For the interval [tex]\( (2, \infty) \)[/tex], choose [tex]\( d = 3 \)[/tex]:
[tex]\[ (3)(3 - 2) = (3)(1) = 3 \geq 0 \][/tex]
This interval satisfies the inequality.
5. Include the critical points [tex]\( d = 0 \)[/tex] and [tex]\( d = 2 \)[/tex] because they make the inequality true:
[tex]\[ 0(0 - 2) = 0 \geq 0 \quad \text{and} \quad 2(2 - 2) = 0 \geq 0 \][/tex]
6. Combine the intervals that satisfy the inequality:
[tex]\[ (-\infty, 0] \cup [2, \infty) \][/tex]
Therefore, the solution set in interval notation is:
[tex]\[ (-\infty, 0] \cup [2, \infty) \][/tex]
This represents all the values of [tex]\( d \)[/tex] that satisfy the inequality [tex]\( d^2 \geq 2d \)[/tex].
1. Rewrite the inequality in standard form:
[tex]\[ d^2 - 2d \geq 0 \][/tex]
2. Factor the left-hand side of the inequality:
[tex]\[ d(d - 2) \geq 0 \][/tex]
3. Identify the critical points by setting each factor equal to zero:
[tex]\[ d = 0 \quad \text{and} \quad d = 2 \][/tex]
These critical points divide the number line into three intervals: [tex]\( (-\infty, 0) \)[/tex], [tex]\( (0, 2) \)[/tex], and [tex]\( (2, \infty) \)[/tex].
4. Test a value in each interval to determine where the product is non-negative:
- For the interval [tex]\( (-\infty, 0) \)[/tex], choose [tex]\( d = -1 \)[/tex]:
[tex]\[ (-1)(-1 - 2) = (-1)(-3) = 3 \geq 0 \][/tex]
This interval satisfies the inequality.
- For the interval [tex]\( (0, 2) \)[/tex], choose [tex]\( d = 1 \)[/tex]:
[tex]\[ (1)(1 - 2) = (1)(-1) = -1 < 0 \][/tex]
This interval does not satisfy the inequality.
- For the interval [tex]\( (2, \infty) \)[/tex], choose [tex]\( d = 3 \)[/tex]:
[tex]\[ (3)(3 - 2) = (3)(1) = 3 \geq 0 \][/tex]
This interval satisfies the inequality.
5. Include the critical points [tex]\( d = 0 \)[/tex] and [tex]\( d = 2 \)[/tex] because they make the inequality true:
[tex]\[ 0(0 - 2) = 0 \geq 0 \quad \text{and} \quad 2(2 - 2) = 0 \geq 0 \][/tex]
6. Combine the intervals that satisfy the inequality:
[tex]\[ (-\infty, 0] \cup [2, \infty) \][/tex]
Therefore, the solution set in interval notation is:
[tex]\[ (-\infty, 0] \cup [2, \infty) \][/tex]
This represents all the values of [tex]\( d \)[/tex] that satisfy the inequality [tex]\( d^2 \geq 2d \)[/tex].
