PreCalculus

7.5. Solved problems

Solve: x + 4 < 1 - 5x
```
x + 4 - 1 + 5x < 0
6x + 3 < 0
2x + 1 < 0
x < -1/2
```
The solution interval is (-∞, -1/2)
Solve: -2 < x + 4 ≤ 8
```
-2 - 4 < x + 4 - 4 ≤ 8 - 4
-6 < x ≤ 4
```
The solution interval is (-6, 4]

Solve: x - 2 ≤ 2x + 3 ≤ x + 4

x - 2 - x - 3 ≤ 2x + 3 - x - 3 ≤ x + 4 - x - 3
   -5 ≤ x ≤ 1
The solution interval is [-5, 1]

Solve: x³ - 6x² + 8x ≤ 0

x(x - 2)(x - 4) ≤ 0

The critical points are x = 0, 2, 4

 x         -  |     +     |     +     |  +   
(x - 2)    -  |     -     |     +     |  +   
(x - 4)    -  |     -     |     -     |  +   
Product    -  |     +     |     -     |  +   
       -------+-----------+-----------+------
              0     1     2     3     4

The inequality is true in the intervals (-∞, 0], [2, 4]

Solve: 2(x - 2) ≤ 5(x + 1) - 12
```
2x - 4 ≤ 5x - 7
2x + 3 ≤ 5x
3 ≤ 3x
1 ≤ x
```
The inequality is true in the interval [1, ∞)
Solve: 2(2 - x) + 6 ≤ 4(x - 2)
```
4 - 2x + 6 ≤ 4x - 8
10 ≤ 6x - 8
18 ≤ 6x
3 ≤ x
```
The inequality is true in the interval [3, ∞)
Solve: -6 ≤ 2x - 10 < 8
```
-3 ≤ x - 5 < 4
 2 ≤ x < 9
```
The inequality is true in the interval [2, 9)
Solve: -6 < 6 - 3x ≤ 9
```
6 > 3x - 6 ≥ -9
12 > 3x ≥ -3
4 > x ≥ -1
```
The inequality is true in the interval [-1, 4)
Solve: 0 < -11x ≤ 132
```
0 > 11x ≥ -132
0 > x ≥ -12
```
The inequality is true in the interval [-12, 0)

Solve: x² - 10 > 3x

x² - 3x - 10 > 0
(x - 5)(x + 2) > 0

(x - 5)    -  |        -        |  +  
(x + 2)    -  |        +        |  +  
Product    +  |        -        |  +  
         -----+-----------------+------
             -2                 5

The inequality is true in the intervals (-∞, -2), (5, +∞)

Solve: x² - 5x < -6

x² - 5x + 6 < 0
(x - 2)(x - 3) < 0

(x - 3)     -   |      -      |   +  
(x - 2)     -   |      +      |   +  
Product     +   |      -      |   +  
         -------+-------------+-------
                2             3

The inequality is true in the interval (2, 3)

Solve: (x + 1)(x - 2)² ≤ 0

(x - 2)²   +  |         +         |  +
(x + 1)   -  |         +         |  +  
Product   -  |         +         |  +  
        -----+-------------------+------
             -1                   2

The inequality is true in the interval (-∞, -1]. Note that the term (x - 2)² is ≥ 0 for all real values of x.

Solve: x² + x + 1 < 0
```
    -1 ± √(-3)
x = ——————————
        2
```
The inequality has no real solution. For all real values of x, the expression on the left hand side is always positive.

Solve: (x + 1)/(x - 5) > 0

(x - 5)    -  |       -       |  +  
(x + 1)    -  |       +       |  +  
Quotient   +  |       -       |  +  
         -----+---------------+------
             -1               5

The inequality is true in the intervals (-∞, -1), (5, +∞)

Solve: (x² + 4x + 3)/(x - 1) ≤ 0

(x + 3)(x + 1)/(x - 1) ≤ 0

(x - 1)    -  |     -     |     -     |  +   
(x + 1)    -  |     -     |     +     |  +   
(x + 3)    -  |     +     |     +     |  +   
Quotient   -  |     +     |     -     |  +   
       -------+-----------+-----------+------
             -3    -2    -1     0     1

The inequality is true in the intervals (-∞, -3], [-1, 1). When x = 1, the expression is undefined.

Solve: (x² - 9)/(x² - 2x + 1) ≥ 0

(x - 3)(x + 3)/(x - 1)² ≥ 0

(x - 3)   -  |       -       |   -   |  +   
(x - 1)²   +  |       +       |   +   |  +   
(x + 3)   -  |       +       |   +   |  +   
Quotient  +  |       -       |   -   |  +   
       ------+---------------+-------+------
             -3               1       3      
The inequality is true in the intervals (-∞, -3], [3, 1). When x = 1, the expression is undefined; however, this point is outside the valid range of the inequality.

Solve:

x + 1
————— ≤ 2
x + 2

Rewrite so that zero is on right hand side:

x + 1       x + 1 - 2x - 4     x + 3
————— - 2 = —————————————— = - ————— ≤ 0
x + 2           x + 2          x + 2

x + 3
————— ≥ 0
x + 2

(x + 2)     -   |      -      |   +  
(x + 3)     -   |      +      |   +  
Product     +   |      -      |   +  
         -------+-------------+-------
               -3            -2

The inequality is true in the interval (-∞, -3], (-2, +∞). When x = -2, the expression is undefined.

Solve:

x - 2
————— > x - 2
  x

Rewrite so that zero is on right hand side:

x - 2             x - 2 - x(x - 2)   x - 2 - x² + 2x
————— - (x - 2) = ———————————————— = ——————————————— > 0
  x                      x                  x

x² - 3x + 2
——————————— < 0
     x

(x - 1)(x - 2)
—————————————— < 0
       x

(x - 2)    -  |     -     |     -     |  +   
(x - 1)    -  |     -     |     +     |  +   
x          -  |     +     |     +     |  +   
Quotient   -  |     +     |     -     |  +   
       -------+-----------+-----------+------
              0           1           2

The inequality is true in the intervals (-∞, 0), (1, 2). When x = 0, the expression is undefined.

Solve: x/(x² + 3x - 10) ≥ 0

x/(x - 2)(x + 5) ≥ 0

(x - 2)    -  |       -       |   -   |  +   
 x         -  |       -       |   +   |  +   
(x + 5)    -  |       +       |   +   |  +   
Quotient   -  |       +       |   -   |  +   
        ------+---------------+-------+------
             -5               0       2

The inequality is true in the intervals (-5, 0], (2, +∞). When x = -5 or 2, the expression is undefined.

Solve:

x + 2     1
————— ≤ —————
  2     x + 1

x + 2    -1     (x + 2)(x + 1) - 2*1
————— + ————— = ———————————————————— ≤ 0
  2     x + 1         2(x + 1)

x² + x + 2x + 2 - 2    x² + 3x
——————————————————— = ———————— ≤ 0
      2(x + 1)        2(x + 1)

x(x + 3)
———————— ≤ 0
2(x + 1)

 x         -  |       -       |   -   |  +   
(x + 1)    -  |       -       |   +   |  +   
(x + 3)    -  |       +       |   +   |  +   
Quotient   -  |       +       |   -   |  +   
        ------+---------------+-------+------
             -3              -1       0

The inequality is true in the intervals (-∞, -3], (-1, 0]. When x = -1, the expression is undefined.