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Absolute Value and Radical Expressions, Equations and Functions

Begin Exam.

Directions: In each of the following exercises, circle the “best” answer
on the exam. Each problem is worth 4 points.

1. Write interval notation for {x| − 7 ≤ x < 2}.

(a) (−7, 2] (b) [−7, 2) (c) [−7, 2]
(d) (−7, 2) (e) (−∞, 2)

2. Find the solutions set for 3(6 − x) < 3x + 2.

(a) (−∞, 10/3) (b) (2/3,+∞) (c) (−∞, 8/3)
(d) (8/3,+∞) (e) (−∞, 2/3)

3. Simplify

(a) a (b) −a (c) |a| (d) 1/a (e) −1/a

4. Find the domain of the function

(a) {x|x ≥ 3} (b) {x|x ≤ 3} (c) {x|x < 3}
(d) {x|x ≥ 0} (e) all real numbers

5. Simplify

(a) x − 5 (b) |x| − 5 (c) |x − 5|
(d) |x + 5| (e)

6. Approximate to three decimal places.

(a) 2469 (b) 111.108 (c) 7.536 (d) 5.927 (e) 6.581

7. Simplify:

8. Simplify:

9. Subtract and simplify:

Directions: Place the solution to each of the following exercises on
your own paper. No more than two problems on one side of a page.
For full credit your work must support your answer. Each problem is
worth 10 points.

Exercise 1. Solve the inequality. Graph the solution on a number
line and express the solution using set-builder notation.
x − 2 <4 or x + 3 ≥ 7

Exercise 2. Solve the inequality with absolute value. Graph the
solution on a number line and express the solution using interval notation.
| − 4x + 5| ≤ 2

Exercise 3. Graph the system of linear inequalities shading the appropriate
region. Label any vertices of the region with their coordi-
nates. For full credit you must show your work for finding the vertices.
x + y ≤ 2
x − y ≥ 4
x ≥ −1

Exercise 4.

(a) Solve |2x + 5| = 3

(b) Rationalize the denominator and simplify:

Exercise 5. Solve:

Exercise 6. Solve:

Solutions to Quizzes

Solution to Question 1: The graph of the inequality −7 ≤ x < 2
is given by

Interval notation for the inequality {x| − 7 ≤ x < 2} is [−7, 2).

Solution to Question 2:

The graph of the inequality x > 8/3 is given by

In interval notation the solution is (8/3,+∞).

Solution to Question 3: Since the index is even simplifies to
|a|.

Solution to Question 4: Given the function is
defined for all x such that 4x − 12 ≥ 0. Solving for x yields x ≥ 3.
The domain of f is {x|x ≥ 3}

Solution to Question 5: Begin simplifying by factoring under the
radical.

Solution to Question 6: Use your calculator to find
6.581. Fractional exponents can also be used,

Solution to Question 7: Simplify by pulling perfect squares from
the square root.

The absolute value is not needed in the denominator since y⁴ > 0 for
all y−values.

Solution to Question 8: Use fractional exponents to simplify.

Solution to Question 9: Simplify the radicals before combining like
terms.

Solutions to Exercises

Exercise 1. First solve each inequality separately.
x − 2 <4 or x + 3 ≥ 7
x <6 or x ≥ 4

Graph each inequality and combine the inequalities for the “or” statement

The last number line shows the solution is all real numbers. In set-builder
notation the solution is x|−∞ < x < ∞.
Exercise 1

Exercise 2. First rewrite the “less than” inequality with absolute
value as an “and” statement without absolute value. Then solve the
inequality.
| − 4x + 5| ≤ 2 => −2 ≤ −4x + 5 ≤ 2.

Now solving yields

−2≤ −4x + 5 ≤ 2
−7≤ −4x ≤ −3
−7/ − 4 ≥ x ≥ −3/ − 4
3/4 ≤ x ≤ 7/4

The graph of the inequality 3/4 ≤ x ≤ 7/4 is given by

In interval notation the solution is [3/4, 7/4]. Exercise 2

Exercise 3. Begin by finding the x−intercepts and y−intercepts for
all of the lines. Then choose a test point, (0, 0), to determine the
region to be shaded.

	x−intercept	y−intercept	test (0, 0)
x + y ≤ 2 x − y ≥ 4 x ≥ −1	(2, 0) (4, 0) (−1, 0)	(0, 2) (0,−4) none	0 + 0 ≤ 2, Yes 0 − 0 ≥ 4, No 0 ≥ −1, Yes

Plotting the intercepts and shading the appropriate regions gives the
following graph.

It remains to find the vertices for the shaded region. Notice there are
only two vertices. One from the intersection of the lines x + y = 2
and x − y = 4. The other intersection is from the line x − y = 4 and
the vertical line x = −1.

To solve the system

Add E1 and E2 to eliminate y. This give
2x = 6
x = 3

Now substitute into E1 and solve for y to get
3 + y = 2
y = −1
One of the vertices is at the point (3,−1).

The second vertex is a simple substitution of x = −1 into the
equation x − y = 4 to give
−1 − y = 4
−y = 5
y = −5

The other vertex is at (−1,−5). Exercise 3

Exercise 4(a) Rewrite the equation with absolute value as two equations.

2x + 5 =−3 or 2x + 5 = 3
2x = −8 or 2x = −2
x = −4 or x = −1

Exercise 4(b) Rationalize the denominator by multiplying both numerator
and denominator by the conjugate of the denominator.

Exercise 5. Remove the radical by squaring both sides.

Setting each factor equal to zero and solving gives
x = 4 and x = −3

Now must check each solution in the original equation.
For x = 4 the left-hand side of the equation is
and the right-hand side of the equation is (4) − 3 = 1. So
x = 4 is a solution.

For x = −3 the left-hand side of the equation is

and the right-hand side of the equation is (−3)−
3 = −6. Since the left-hand and right-hand side of the equations do
not agree, x = −3 is not a solution. Exercise 5

Exercise 6. Isolate one of the radicals and square both sides to
remove the radical. Repeat this procedure until all of the radicals are
gone.

Check the solution by substituting x = 4 into the original equation.
The left-hand side of the equation becomes
which is also the right-hand side of the equation. So x = 4 is the
solution. Exercise 6