|a − b| = |b − a| for and real numbers a and b . True. We’ve seen that a − b = −(b − a), i.e. a-b and b-a are
which means their absolute values are the same.
The graph of a rational function can cross it’s
vertical asymptote. False. A rational function f has a vertical asymptote at x = a if
f(a) is undefined because of division by zero. Therefore the point (a,
cannot be on the graph.
The domain of any rational function is all real
False. Let f(x) =1/x, the domain is x ≠ 0.
If x > 2 or x < −3, then we can say 2 < x < −3. False. 2 < x < −3 means 2 < x AND x < −3.
There are no numbers that are more than 2 and less than -3 whereas
5 (for example) is a number that is more than 2 or less than -3
True. The common denominator is ab hence
1. Solve the fol lowing equations for x. Where applicable,
write the solution sets in interval
This inequality is saying that the number 2x + 3 is more
than 15 units away from 0.
Hence we can rewrite as
Solving each equation we get that
The union of these two solution
sets in interval notation is
(b) |2x + 1| = 7
This means that the number 2x + 1 is 7 units away from 0
2x + 1 = 7
2x + 1 = −7
Solving each equation we see that x = 3 or x = −4.
Remember that for absolute
value equations we should always check our solutions in the original equation.
(c) 3x ≤ 15 and −2x < 6
We need to solve each equation and then take the intersection of both solution
The first inequality gives us x ≤ 5 and the second
inequality gives us x > −3, (remember
to switch the direction of the inequality when you divide by a negative number ).
Numbers that are greater than −3 and less than or equal to
5 are in the interval
(d) (x − 1)(x + 4) = 14
Remember that the first step is to put the equation in
standard form, everything on
one side of the of the equal sign and 0 on the other:
Using the zero product principle the last line implies
that x = −6 or x = 3.
8x3 + 27 is a sum of cubes : (2x)3 + 33.
x4 − 1 is a difference of squares: (x2)2 − 1.
x4 + x2 − 2 is a closet quadratic; let t = x2 and factor t2 + t − 2.
4x2 + 12x + 9 is a perfect square : (2x)2 + 2(2x)(3) + 32.
To subtract we need to find the least common denominator.
To do this we need
to completely factor each denominator:
therefore the least common denominator is (x − 2)(x + 1)(x
3. Consider the graph of the rational function f below:
(a) What is f(1)?
f(1) = −1
(b) What is the domain of f?
Remember that the domain is the set of all possible inputs, all the values for x
the graph lies above, below or on the x−axis.
So the domain is x ≠ 2 or x ≠ −2. In interval notation:
(c) What is the range of f?
The range is the set of outputs, or all values for y where the graph lies to the
left or on the y−axis.
So the range is
(d) What are the vertical asymptotes of f?
x = 2 and x = −2
(e) When x gets very very large, what can you say about
f(x) gets closer and closer to 3.
4. The function
models the number of people, f(x), in millions, receiving
food stamps x years after 1990.
(a) How many people received food stamps in 1996?
x is the number of years after 1990, so compute f(6):
So 26 million people received food stamps in 1996.
(b) In which years did 25 million peoples receive food
Now we know how many people received food stamps, we want to determine in what
year, i.e. we know our output, we need to find the input. So solve for x in the
We begin by putting the equation into standard form.
move 25 to the other side
multiply everything by -4 to clear the
So x = 8 or x = 4 which means that 25 million people
received food stamps in the years
1994 and 1998.
5. What is the domain of
? State your answer any way you’ d like
notation, set notation, or short-hand).
To find the domain we just need to figure out what values
for x would make the denominator
0, which means we need to solve the equation
, which means we need to factor 4x^2 − 13x + 3.
Since 4x^2 − 13x + 3 isn’t a perfect square probably the
“ac” method is best :
ac = 12
(−12)(−1) = 12 and −12 + (−1) = −13
Hence the solutions to the equation 4x^2 − 13x + 3 = 0 are
x = 3 or x =1/4
and thus the domain of f is
7. The x-intercept(s) of the function y = x^2 + 6x + 8
To find the x-intercepts just find the solutions to the equation x^2 + 6x + 8 =
so the x-intercepts are −4 and −2.
Extra Credit: At the end of the day, the change
machine at a laundrette contained at least
$3.20 and at most $5.45 in nickels, dimes, and quarters. There were 3 fewer
dimes than twice
the number of nickels and 2 more quarters than twice the number of nickels. What
was the least
possible number and the greatest possible number of nickels?
HINT Let your unknown be the number of nickels and
write the number of quarters and dimes in
terms of that. Instead of worrying about arithmetic withdecimals just remember
is in terms of dollars and cents, you know how to add money , I know you do!
Let n be the number of nickels so the number of dimes is
2n − 3 and the number of quarters
is 2n + 2.
The total value of the change is 0.05n + 0.10(2n − 3) + 0.25(2n + 2).
We are given that the value of the change is at least $3.20 and at most $5.45
To solve for n, rather than worry about dividing by
decimals, remember that this is just all
money. So 3.00/0.75 is just however many times 75 cents are in $3.00, 4 and
5.25/0.75 is just however
many times 75 cents are in $5.25, 7. So