Question

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Find the solution set for each inequality.

Answer 1

41. (-∞, -1] ∪ [2, 3)

42. (2, 5) ∪ (5, ∞)

Explanation:

Question 41

Given rational inequality:

`(x^2-x-2)/(x-3) \leq 0`

Factor the numerator:

`((x-2)(x+1))/(x-3) \leq 0`

Find the roots by solving f(x) = 0 (set the numerator to zero):

`\implies (x-2)(x+1)=0`

`\implies x-2=0 \implies x=2`

`\implies x+1=0 \implies x=-1`

Find the restrictions by solving f(x) = undefined (set the denominator to zero):

`\implies x-3=0 \implies x=3`

Create a sign chart, using closed dots for the roots and open dots for the restrictions (see attachment 1).

Choose a test value for each region, including one to the left of all the critical values and one to the right of all the critical values.

• Test values: -2, 0, 2.5, 4

For each test value, determine if the function is positive or negative:

`f(-2)=((-2)^2-(-2)-2)/((-2)-3) =(+)/(-)=-`

`f(0)=((0)^2-(0)-2)/((0)-3) =(-)/(-)=+`

`f(2.5)=((2.5)^2-(2.5)-2)/((2.5)-3) =(+)/(-)=-`

`f(4)=((4)^2-(4)-2)/((4)-3) =(+)/(+)=+`

Record the results on the sign chart for each region (see attachment 1).

As we need to find the values for which f(x) ≤ 0, shade the appropriate regions (zero or negative) on the sign chart (see attached).

Therefore, the solution set is:

• x ≤ -1 or 2 ≤ x < 3

As interval notation:

• (-∞, -1] ∪ [2, 3)

Question 42

Given rational inequality:

`(√(x))/((x-2)|x-5|) > 0`

Find the roots by solving f(x) = 0 (set the numerator to zero):

`\implies √(x)=0 \implies x=0`

Find the restrictions by solving f(x) = undefined (set the denominator to zero):

`\implies (x-2)|x-5|=0`

`\implies x-2=0 \implies x=2`

`\implies |x-5|=0 \implies x=5`

Create a sign chart, using closed dots for the roots and open dots for the restrictions (see attachment 1).

Choose a test value for each region, including one to the left of all the critical values and one to the right of all the critical values.

• Test values: -1, 1, 3, 6

For each test value, determine if the function is positive or negative:

`f(-1)=(√(-1))/((-1-2)|-1-5|)=(\sf unde\:\!fined)/(-)=\sf unde\:\!fined`

`f(1)=(√(1))/((1-2)|1-5|)=(+)/(-)=-`

`f(3)=(√(3))/((3-2)|3-5|)=(+)/(+)=+`

`f(6)=(√(6))/((6-2)|6-5|)=(+)/(+)=+`

Record the results on the sign chart for each region (see attachment 2).

As we need to find the values for which f(x) > 0, shade the appropriate regions (positive) on the sign chart (see attached).

Therefore, the solution set is:

• 2 < x < 5 or x > 5

As interval notation:

• (2, 5) ∪ (5, ∞)