CHAPTER 1 (Applied exercises)
What’s the domain of each of the following functions?
1
1) 𝑓(𝑥) = 𝑥−2
Notice that there is no output number when the input number equals 2, since division by
zero is meaningless. Thus x can have any value except 2. We say that 2 is not in the domain
of the function f, or, in intervals notation (−∞, 2[ ∪ ]2, ∞)
2) 𝑔(𝑥) = √𝑥 + 5
Since the quantity under the radical must be non-negative, it follows that x+5 ≥ 0. Thus the
implicit domain is the interval [-5, ∞).
1
3) ℎ(𝑥) = 𝑥√𝑥−1
Since we cannot divide by 0 we know that x cannot equal to either 0 or 1. And since we
cannot take the square root of a negative number (recall that output numbers must be real
numbers), then x−1≥0. Taken together these two restrictions mean that x−1>0 or x>1. Thus
the implicit domain is the interval ]1, ∞)
What’s the composite function f(g(x))?
1) 𝑓(𝑥) = 3𝑥² − 3 and 𝑔(𝑥) = 1 − 𝑥
𝑓(𝑔(𝑥)) = 𝑓(1 − 𝑥) = 3(1 − 𝑥 2 ) − 3 = 3𝑥² − 6𝑥 , for all x
2) 𝑓(𝑥) = √4 − 𝑥 and 𝑔(𝑥) = √𝑥
𝑓(𝑔(𝑥)) = 𝑓(√𝑥) = √4 − √𝑥 ; for x ≥ 0 and 4 − √𝑥 ≥ 0. Therefore, 0 ≤ 𝑥 ≤ 16. In interval
notation, the domain of the composite function f(g(x)) is [0, 16].
Find the equation of the line containing the points (−3,5) and (0,7)
Since the line contains the points (−3,5) and (0,7) the slope is:
𝑚=
𝑦₂ − 𝑦₁ 7 − 5 2
=
=
𝑥₂ − 𝑥₁ 0— 3 3
Considering the point (-3,5) and knowing that the equation of the line is of the form: 𝑦 =
2
3
𝑥 + 𝑏 we just substitute x by -3 and y by 5 to find b = 7.
2
So the equation of the line is 𝑦 = 3 𝑥 + 7
Find the point-slope form and line equation of the line with slope 3 containing
the point (2,−3).
The line has slope m=3 and contains the point (x1, y1) = (2,−3)
So the point-slope formula y − y1 = m(x − x1) gives the result: y + 3 = 3(x − 2).
If we solve this equation for y we get the line equation which is: y = 3x − 9.
RECALL!!!!!
The quadratic formula:
If 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0, then 𝑥 =
−𝑏±√𝑏2 −4𝑎𝑐
2𝑎
The expression ∆= 𝑏² − 4𝑎𝑐 under the radical is called the discriminant.
Some important things to know about the discriminant:
1. 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 has only one solution if and only if the discriminant equals zero.
2. 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 has two real solutions if and only if the discriminant is positive.
3. 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 has no solution if and only if the discriminant is negative.
Solve for x: 2𝒙² + 𝟖𝒙 + 𝟏 = 0
In this case a = 2, b = 8, and c = 1. Using the discriminant as we’ve seen above:
∆= 𝑏² − 4𝑎𝑐 = 8² − 4(2)( 1) = 56
Since ∆ > 0, the quadratic equation has two solutions, and they are
−𝑏 ± √𝑏 2 − 4𝑎𝑐 −8 ± √56
√14
𝑥=
=
= −2 ± (
)
2𝑎
4
2
What is the domain of? 𝒇(𝒙) =
𝟒𝒙𝟐 −𝟑
𝟐𝒙𝟐 +𝒙−𝟑
Since division by zero is not allowed, the domain excludes all the values of x that are the
solutions (if they exist) of the quadratic equation: 2x² + x – 3 = 0
For this equation: a = 2, b = 1, and c = -3. Using the discriminant:
∆= 𝑏 2 − 4𝑎𝑐 = 12 − 4(2)(−3) = 25
Since ∆ > 0, the quadratic equation has two solutions, and they are
𝑥1 =
−𝑏+√∆
2𝑎
=
−1+√25
4
=1
𝑥2 =
And
3
3
2
2
−𝑏−√∆
2𝑎
=
−1−√25
4
=−
3
2
So the domain of the function is: (−∞,− [ ∪ ] − , 1[ ∪ ]1, ∞)
What’s the distance AB, and the line equation AB knowing that A(1, 2) and
B(4, 6) ?
DAB = √(𝑥₂ − 𝑥₁)² + (𝑦₂ − 𝑦₁)² = √(4 − 1)² + (6 − 2)² = 5
To find the equation AB, let’s begin first by finding the slope:
𝑚=
𝑦₂ − 𝑦₁ 6 − 2 4
=
=
𝑥₂ − 𝑥₁ 4— 1 3
4
Now, we know that LAB = 𝑦 = 3 𝑥 + 𝑏
2
And if we consider point A and substitute by its coordinates to solve for b, we find that b = 3
4
2
So LAB equation is: 𝑦 = 3 𝑥 + 3
Sketch an approximate graph of 𝒇(𝒙) = 𝟐𝒙² + 𝒙 − 𝟑 and find its intersection
with the line 𝑦 = 3𝑥 – 3?
𝑏
1
Recall that the abscissa of the vertex is: Xvertex – 2𝑎 = − 4 substituting in f(x) we find that
25
Yvertex = − 8 , we know now the exact location of the vertex.
The parabola opens upward since a > 0.
3
X-intercepts: we make y = 0 and solve the quadratic equation to get X1 = 1 and X2 = − 2
Y-intercept: we just make X = 0 to find that Y = -3
The intersection of the two graphs occurs when they have the abscissa and ordinate, so we
can just solve for: 2𝑥² + 𝑥 − 3 = 3𝑥 − 3 solving for x gives two solutions which are 0
and 1, that we substitute in one of the functions to find Y. so the points of intersection are
(0, -3) and (1, 0). Below is the graph of the two functions:
Wording problem:
Since the beginning of the year, the price of a bottle of soda at a local discount supermarket
has been rising at a constant rate of 2 cents per month. By November first, the price had
reached $1.56 per bottle. Express the price of the soda as a function of time and determine
the price at the beginning of the year.
Let x denote the number of months that have elapsed since the first of the year and y the
price of a bottle of soda (in cents). Since y changes at a constant rate with respect to x, the
function relating y to x must be linear, and its graph is a straight line. Since the price y
increases by 2 each time x increases by I, the slope of the line must be 2. The fact that the
price was 156 cents ($1.56) on November first, 10 months after the first of the year, implies
that the line passes through the point (10, 156). To write an equation defining y as a function
of x, use the point-slope formula
Y – Y0 = m(X – X0) with m = 2, X0 = 10, and Y0 = 156 to get: Y – 156 = 2(X – 10)
Or y = 2x + 136
Find the slope and y-intercept of the line L2 that is perpendicular to L1: y=3x+4
and passes through the point (3, −3)?
By definition, any line perpendicular to L1 has a slope of -1/3 since the slope of L1 is 3.
1
So the general form of L2 equation is: y = − 3 x + b
𝟏
Substituting (x, y) by (3, -3) gives b = -2  L2: 𝐲 = − 𝟑 𝐱 − 𝟐
What do the following two equations represent?
𝟒𝒙 + 𝟑𝒚 = 𝟒 And 𝟖𝒙 + 𝟔𝒚 = 𝟒
𝟒
𝟒
𝟑
𝟑
The two equations can be written as: 𝐲 = − 𝐱 +
𝟒
𝟐
𝟑
𝟑
and 𝐲 = − 𝐱 +
Since they have the same slope and they are equations of lines, we can say that the
equations represent two parallel lines.
CAR RENTAL
A car rental agency charges $75 per day plus 70 cents per mile.
a. Express the cost of renting a car from this agency for 1 day as a function of the
number of miles driven and draw the graph.
b. How much does it cost to rent a car for a 1-day trip of 50 miles?
c. The agency also offers a rental for a flat fee of $125 per day. How many miles
must you drive on a 1-day trip for this to be the better deal?
a. 𝑓(𝑥) = 0.7𝑥 + 75 where x is the number of miles driven and f is expressed in
dollars
b. It will cost 0.7 * 50 + 75 = $110
c. Let’s first find when the first option is equal to $125 by solving for x: 125 = 0.7𝑥 + 75
For this case: x = 50/0.7 ≈ 71 Miles. This informs us that below 71 miles it’s better to
choose the first option, and above 71 miles it’s better to choose the $125 deal. The
following graph will help you better understand:
Find the slope, x and y intercept of the line 3y +2x = 6 and draw the graph?
First put the equation 3y + 2x = 6 into slope-intercept form y = mx + b
To do this, solve for y to get:
3y = -2x + 6  𝑦 = −2/3 𝑥 + 2
It follows that the slope is -2/3, the y-intercept is 2, and consequently, the x-intercept is 3.
Find
𝒍𝒊𝒎 ( 𝟑𝒙𝟑 − 𝟒𝒙 + 𝟖) and lim
3𝑥 3 −8
𝑥→1 𝑥−2
𝒙→−𝟏
?
1. Apply the properties of limits to obtain
lim( 3𝑥 3 − 4𝑥 + 8) = 3( lim 𝑥)3 – 4( lim 𝑥)+ lim 8
𝑥→−1
𝑥→−1
𝑥→−1
𝑥→−1
= 3(-1)3 - 4(-1) + 8 = 9
2. Since lim(𝑥 − 2) ≠ 0, you can use the quotient rule for limits to get:
𝑥→1
𝑙𝑖𝑚
𝑥→1
3𝑥 3 −8
𝑥−2
𝑙𝑖𝑚(3𝑥 3 −8)
3 𝑙𝑖𝑚 𝑥 3 −𝑙𝑖𝑚 8
= 𝑥→1
=
𝑙𝑖𝑚(𝑥−2)
𝑥→1
3−8
= 1−2 = 5
𝑥−1
𝑥−1
𝑙𝑖𝑚 𝑥−𝑙𝑖𝑚 2
𝑥→1
𝑥→1
Find 𝒍𝒊𝒎
𝒙𝟐 −𝟏
?
𝒙→𝟏 𝒙𝟐 −𝟑𝒙+𝟐
As x approaches 1, both the numerator and the denominator approach zero, and you can
draw no conclusion about the size of the quotient. To proceed, observe that the given
function is not defined when x = 1 but that for all other values of x, you can cancel the
common factor x - 1 to obtain
𝑥 2 −1
𝑥 2 −3𝑥+2
(𝑥−1)(𝑥+1)
𝑥+1
= (𝑥−1)(𝑥−2) = 𝑥−2
x≠ 1
(Since x ≠1, you are not dividing by zero.) Now take the limit as x approaches (but is not
equal to) 1 to get:
𝑥 2 −1
lim (𝑥+1)
2
lim 𝑥 2 −3𝑥+2 = 𝑥→1
= = -2
lim 𝑥−2) −1
𝑥→1
𝑥→1(
The graph below shows geometrically what is written above:
What’s the 𝒍𝒊𝒎
𝒙𝟐
𝒙→+∞ 𝟏+𝒙+𝟐𝒙𝟐
?
𝑥2
To get a feeling of what happens with this limit, we can evaluate 𝑓(𝑥) = 1+𝑥+2𝑥 2 as x
increasing without bound. Here’s a table of that
X
F(x)
100
0.49749
1,000
0.49975
10,000
0.49997
100,000
0.49999
The above table suggests that f(x) tends toward 0.5 as x grows larger and larger.
To confirm this observation analytically, we divide each term in f(x) by the highest power
that appears in the denominator 1 + x + 2x²; namely, x². This enables us to find lim 𝑓(𝑥) by
𝑥→+∞
applying reciprocal power rules as follows:
lim
𝑥→+∞
𝑥2
1+𝑥+2𝑥 2
= lim
𝑥 2 /𝑥 2
2
𝑥→+∞ 1 + 𝑥 +2𝑥 2 +2𝑥
𝑥2 𝑥2
𝑥2
=
Find 𝒍𝒊𝒎
lim 1
x→+∞
1
1
lim 2 + lim + lim 2
x→+∞x
x→+∞x x→+∞
−𝒙𝟑 +𝟐𝒙+𝟏
𝒙→+∞
𝒙−𝟑
=
1
0+0+2
= 0.5
?
The highest power in the denominator is x. Divide both numerator and denominator by x to
get:
lim
𝑥→+∞
−𝑥 3 +2𝑥+1
𝑥−3
= lim
𝑥→+∞
1
−𝑥 2 +2+
3
𝑥
1−
Since: lim (−𝑥 2 + 2 + 𝑥) = −∞
𝑥→+∞
It follows that:
1
𝑥
and
𝐥𝐢𝐦
𝒙→+∞
3
lim (1 − 𝑥) = 1
𝑥→+∞
−𝒙𝟑 +𝟐𝒙+𝟏
𝒙−𝟑
=-∞
Determine whether 𝒍𝒊𝒎 𝒇(𝒙)exists, where:
𝒙→𝟏
𝑥 + 1, 𝑖𝑓 𝑥 < 1
𝑓(𝑥) = { 2
−𝑥 + 4𝑥 − 1, 𝑖𝑓 𝑥 ≥ 1
Computing the one-sided limit at x = 1, we find that: 1-
𝒍𝒊𝒎 𝒇(𝒙) = 𝒍𝒊𝒎 (𝒙 + 𝟏) = 1 +1 = 2
𝒙→𝟏−
𝒙→𝟏−
𝒍𝒊𝒎 𝒇(𝒙) = 𝒍𝒊𝒎 (−𝒙𝟐 + 𝟒𝒙 − 𝟏) = - (1²) + 4(1) – 1 = 2
𝒙→𝟏+
𝒙→𝟏+
Since the two one-sided limits are equal, it follows that:
𝒍𝒊𝒎 𝒇(𝒙) = 𝒍𝒊𝒎 𝒇(𝒙) = 𝒍𝒊𝒎 𝒇(𝒙) = 2
𝒙→𝟏
𝒙→𝟏−
𝒙→𝟏+