1/30/24, 8:24 PMPre-Class Assignment Week 4
Pre-Class Assignment Week 4
Due Wednesday by 11:30am
Points 50
Submitting an external tool
Pre-Class Assignment Week 4
You’re encouraged to work on this in the lab.
See the Home Page (https://fiu.instructure.com/courses/186802/pages/welcome-to-college-algebraa-hybrid-learning-course) for specific lab hours.
Be sure to ask for help if you need it!
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Pre-Class Assignment Week 4
Pre-Class Assignment Week 4
ariel gonzalez
Due in 15 hours, 6 minutes. Due Wed 01/31/2024 11:30 am
Read Section 2.2 “More on Graphs” to prepare for class.
In this week’s pre-requisite module, we covered the following topics: topics
evaluating polynomial, rational and radical expressions. These are crucial skills
needed for the conversation that we will be having this week about average rate of
change and piecewise functions. Remember – you can always work on this
assignment in the Mastery Math Lab if you need additional help.
Many times you will come across real-world situations that can’t be
modeled by a function with a single simple formula. Rather, you will need
two or three, or even many different formulas to capture the function.
The different formulas will apply over different portions of the domain.
This is referred to as a “piece-wise” defined function.
QUESTION 1
Calculate the following values, using the given function:
12x + 5
if
x < 0
− 4x + 6
if
0 ≤ x < 9
h(x) = {
A) The value of h( − 6)
is not defined, becuase − 6 is not in the domain of h
is not defined, becuase − 6 is not a rule of h
is equal to 12 ⋅ − 6 + 5 because you use the rule corresponding to − 6 < 0 being true
is equal to − 4 ⋅ − 6 + 6 because you use the rule corresponding to 0 ≤ − 6 < 9 being true
B) h(0) =
C) h(3) =
D) The value of h(14) is
is equal to 12 ⋅ 14 + 5 because you use the rule corresponding to 0 ≤ 14 < 9 being true
is not defined, because 14 is not a rule of h
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Pre-Class Assignment Week 4
is not defined, becuase 14 is not in the domain of h
is equal to − 4 ⋅ 14 + 6 because you use the rule corresponding to 0 ≤ 14 < 9 being true
QUESTION 2
This function is a piece-wise defined function:
11x − 8
⎧
⎪
h(x) = ⎨ x
⎩
⎪
7
2
+ 13
if
x <
− 6
if
− 6 ≤ x < 8
if
x > 8
A) Which expression should we use to evaluate h at x = − 14?
Because − 6 ≤ − 14 < 8 is true, we use the expression x + 13
2
Because − 14 < − 6 is true, we use the expression 11x − 8
Because − 14 > 8 is true, we use the expression 7
B) The value of h at x = − 14 is
C) Which expression should we use to evaluate h at x = 0
Because 0 < − 6 is true, we use the expression 11x − 8
Because − 6 ≤ 0 < 8 is true, we use the expression x + 13
2
Because 0 > 8 is true, we use the expression 7
D) The value of h at x = 0 is
E) What will the y-intercept of the h be?
There is no y-intercept
(13, 0)
(0, 0)
(0, 13)
F) The value of h at x = 11 is
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G) What will the last piece (where h(x) = 7) of the graph look like for this function?
12
11
10
9
8
7
6
5
4
3
2
1
-12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1
-1
1
2
3
4
5
6
7
8
9 10 11 12
1
2
3
4
5
6
7
8
9 10 11 12
1
2
3
4
5
6
7
8
9 10 11 12
-2
-3
-4
-5
-6
-7
-8
-9
-10
-11
-12
12
11
10
9
8
7
6
5
4
3
2
1
-12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1
-1
-2
-3
-4
-5
-6
-7
-8
-9
-10
-11
-12
12
11
10
9
8
7
6
5
4
3
2
1
-12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1
-1
-2
-3
-4
-5
-6
-7
-8
-9
-10
-11
-12
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Pre-Class Assignment Week 4
12
11
10
9
8
7
6
5
4
3
2
1
-12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1
-1
1
2
3
4
5
6
7
8
9 10 11 12
-2
-3
-4
-5
-6
-7
-8
-9
-10
-11
-12
QUESTION 3
We want to define a function that is represented by this graph y = f (x):
10
9
8
7
6
5
4
3
2
1
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1
-1
1
2
3
4
5
6
7
8
9 10
-2
-3
-4
-5
-6
-7
-8
-9
-10
A) If the blue line (using its pattern) continued on forever, the full line would represent the graph of
which equation:
y =
− 2
x =
− 2
y = 3
x = 3
B) If the red line (using it pattern) continuted on forever, the full line would represent the graph of which
equation:
y =
− 5
± 5
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y = x ± 5
y =
− 3x + 1
x = 2
The domain of f is formed by joining the part of the domain that the blue line in the
graph contributes, together with the part of the domain that the red line in the graph
contributes.
C) Look at the graph, what is the domain for the blue line?
D) Look at the graph, what is the domain for the red line?
E) The full definition of the function f is therefore:
⎧
⎪
f (x) = ⎨
⎩
⎪
if x ≤
if x >
The average rate of change of a function from a to b is given by
f (b) − f (a)
b − a
This provides a tool to understand the behavior of a function, by
comparing together the amount of change in the input quantity to the
amount of change in the output quanity. This type of multiplicative
relationship is a preview of several important ideas in Calculus.
QUESTION 4
Consider the function h(x) = 3x + 13
A) What is the average rate of change of h from 17 to 23?
B) What is the average rate of change of h from 4 to 20?
C) What do you notice from a) and b)?
the average rate of change outcomes are different, one negative and one positive
the average rate of change outcomes don’t exist because of division by 0 in the fractions
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the average rate of change outcomes don t exist because of division by 0 in the fractions
the average rate of change outcomes are the same
the average rate of change outcomes are different, but both positive values
D) What type of a function is f ?
a constant function
a quadratic function
a linear function
a piecewise function
E) What is the slope of the line y = 3x + 13?
F) For a linear function, the average rate of change must be the same as the
y
-intercept
slope
x
-intercept
of the line.
QUESTION 5
Consider the function h(x) = 11x + 18x − 6
2
A) What is the average rate of change of h from 13 to 17?
B) What is the average rate of change of h from 10 to 12?
C) What type of function is h?
a constant function
a quadratic function
a linear function
a piecewise function
D) Is the average rate of change constant or varying for h?
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constant, because for every polynomial function the average rate of change function must be
constant
varying, because for every function the average rate of change quantity must vary
varying, because A) and B) have different outcomes
constant, because for every function the average rate of change quantity is constant
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