Math / Discrete

Question12. The sequence described by the explicit rule: $f(n)=42-7(n-1)$ \begin{tabular}{l|l} $n$ & $f(x)$ \\ \hline 1 & 42 \\ 2 & 35 \\ 3 & 28 \\ 4 & 21 \end{tabular}
First five terms of the sequence: Domain: $\{1,2,3,4\}$ Range: x - Intercept(s): y - Intercept(s): Interval(s) of increase: non- Interval(s) of decrease: Discrete Continuous/Discontinuous:

Studdy Solution

STEP 1

What is this asking? We're given a formula $f(n) = 42 - 7(n-1)$ that generates a sequence of numbers, and we need to find the first five terms, the domain and range for the first four terms, the x and y intercepts, where it increases or decreases, and whether it's discrete or continuous. Watch out! Remember, $n$ starts at 1, not 0!
Also, don't mix up domain and range, and think carefully about what "continuous" and "discrete" mean for a sequence.

STEP 2

1. Calculate the first five terms
2. Determine the domain and range
3. Find the x and y intercepts
4. Identify intervals of increase or decrease
5. Determine if the sequence is discrete or continuous

STEP 3

Let's calculate the first five terms!
We'll plug in $n = 1, 2, 3, 4, 5$ into our formula $f(n) = 42 - 7(n-1)$ .

STEP 4

For $n = 1$ , $f(1) = 42 - 7(1-1) = 42 - 7 \cdot 0 = 42 - 0 = 42$ .
Our first term is 42!

STEP 5

For $n = 2$ , $f(2) = 42 - 7(2-1) = 42 - 7 \cdot 1 = 42 - 7 = 35$ .
Our second term is 35!

STEP 6

For $n = 3$ , $f(3) = 42 - 7(3-1) = 42 - 7 \cdot 2 = 42 - 14 = 28$ . 28 is our third term!

STEP 7

For $n = 4$ , $f(4) = 42 - 7(4-1) = 42 - 7 \cdot 3 = 42 - 21 = 21$ .
The fourth term is 21!

STEP 8

For $n = 5$ , $f(5) = 42 - 7(5-1) = 42 - 7 \cdot 4 = 42 - 28 = 14$ .
And finally, our fifth term is 14!

STEP 9

The domain is the set of inputs, which are the first four $n$ values: $\{1, 2, 3, 4\}$ .

STEP 10

The range is the set of outputs, which are the first four terms we calculated: $\{42, 35, 28, 21\}$ .

STEP 11

The y-intercept is the value of $f(n)$ when $n = 0$ .
However, since our sequence starts at $n = 1$ , there's no y-intercept.

STEP 12

The x-intercept is the value of $n$ when $f(n) = 0$ .
So, we set $0 = 42 - 7(n-1)$ and solve for $n$ .

STEP 13

$0 = 42 - 7(n-1)$ .
Subtract 42 from both sides: $-42 = -7(n-1)$ .

STEP 14

Divide both sides by -7: $6 = n - 1$ .

STEP 15

Add 1 to both sides: $n = 7$ .
So, the x-intercept is 7.

STEP 16

Notice that as $n$ increases, $f(n)$ decreases.
This means the sequence is decreasing.

STEP 17

Since the sequence is defined only for integer values of $n$ , it's a discrete sequence.
We can't have a term between, say, the first and second term.

STEP 18

First five terms: 42, 35, 28, 21, 14 Domain: {1, 2, 3, 4} Range: {42, 35, 28, 21} x-intercept: 7 y-intercept: None Intervals of increase: None Intervals of decrease: All Discrete/Continuous: Discrete

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Studdy Solution

STEP 1

STEP 2

1. Calculate the first five terms
2. Determine the domain and range
3. Find the x and y intercepts
4. Identify intervals of increase or decrease
5. Determine if the sequence is discrete or continuous

STEP 3

Let's calculate the first five terms!
We'll plug in $n = 1, 2, 3, 4, 5$ into our formula $f(n) = 42 - 7(n-1)$ .

STEP 4

For $n = 1$ , $f(1) = 42 - 7(1-1) = 42 - 7 \cdot 0 = 42 - 0 = 42$ .
Our first term is 42!

STEP 5

For $n = 2$ , $f(2) = 42 - 7(2-1) = 42 - 7 \cdot 1 = 42 - 7 = 35$ .
Our second term is 35!

STEP 6

For $n = 3$ , $f(3) = 42 - 7(3-1) = 42 - 7 \cdot 2 = 42 - 14 = 28$ . 28 is our third term!

STEP 7

For $n = 4$ , $f(4) = 42 - 7(4-1) = 42 - 7 \cdot 3 = 42 - 21 = 21$ .
The fourth term is 21!

STEP 8

For $n = 5$ , $f(5) = 42 - 7(5-1) = 42 - 7 \cdot 4 = 42 - 28 = 14$ .
And finally, our fifth term is 14!

STEP 9

The domain is the set of inputs, which are the first four $n$ values: $\{1, 2, 3, 4\}$ .

STEP 10

The range is the set of outputs, which are the first four terms we calculated: $\{42, 35, 28, 21\}$ .

STEP 11

The y-intercept is the value of $f(n)$ when $n = 0$ .
However, since our sequence starts at $n = 1$ , there's no y-intercept.

STEP 12

The x-intercept is the value of $n$ when $f(n) = 0$ .
So, we set $0 = 42 - 7(n-1)$ and solve for $n$ .

STEP 13

$0 = 42 - 7(n-1)$ .
Subtract 42 from both sides: $-42 = -7(n-1)$ .

STEP 14

Divide both sides by -7: $6 = n - 1$ .

STEP 15

Add 1 to both sides: $n = 7$ .
So, the x-intercept is 7.

STEP 16

Notice that as $n$ increases, $f(n)$ decreases.
This means the sequence is decreasing.

STEP 17

Since the sequence is defined only for integer values of $n$ , it's a discrete sequence.
We can't have a term between, say, the first and second term.

STEP 18

First five terms: 42, 35, 28, 21, 14 Domain: {1, 2, 3, 4} Range: {42, 35, 28, 21} x-intercept: 7 y-intercept: None Intervals of increase: None Intervals of decrease: All Discrete/Continuous: Discrete

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Studdy Solution

STEP 1

STEP 2

1. Calculate the first five terms2. Determine the domain and range3. Find the x and y intercepts4. Identify intervals of increase or decrease5. Determine if the sequence is discrete or continuous

STEP 3

Let's **calculate** the first five terms!We'll plug in n=1,2,3,4,5n = 1, 2, 3, 4, 5n=1,2,3,4,5 into our formula f(n)=42−7(n−1)f(n) = 42 - 7(n-1)f(n)=42−7(n−1).

STEP 4

For n=1n = 1n=1, f(1)=42−7(1−1)=42−7⋅0=42−0=42f(1) = 42 - 7(1-1) = 42 - 7 \cdot 0 = 42 - 0 = 42f(1)=42−7(1−1)=42−7⋅0=42−0=42.Our **first term** is **42**!

STEP 5

For n=2n = 2n=2, f(2)=42−7(2−1)=42−7⋅1=42−7=35f(2) = 42 - 7(2-1) = 42 - 7 \cdot 1 = 42 - 7 = 35f(2)=42−7(2−1)=42−7⋅1=42−7=35.Our **second term** is **35**!

STEP 6

For n=3n = 3n=3, f(3)=42−7(3−1)=42−7⋅2=42−14=28f(3) = 42 - 7(3-1) = 42 - 7 \cdot 2 = 42 - 14 = 28f(3)=42−7(3−1)=42−7⋅2=42−14=28. **28** is our **third term**!

STEP 7

For n=4n = 4n=4, f(4)=42−7(4−1)=42−7⋅3=42−21=21f(4) = 42 - 7(4-1) = 42 - 7 \cdot 3 = 42 - 21 = 21f(4)=42−7(4−1)=42−7⋅3=42−21=21.The **fourth term** is **21**!

STEP 8

For n=5n = 5n=5, f(5)=42−7(5−1)=42−7⋅4=42−28=14f(5) = 42 - 7(5-1) = 42 - 7 \cdot 4 = 42 - 28 = 14f(5)=42−7(5−1)=42−7⋅4=42−28=14.And finally, our **fifth term** is **14**!

STEP 9

The **domain** is the set of inputs, which are the first four nnn values: {1,2,3,4}\{1, 2, 3, 4\}{1,2,3,4}.

STEP 10

The **range** is the set of outputs, which are the first four terms we calculated: {42,35,28,21}\{42, 35, 28, 21\}{42,35,28,21}.

STEP 11

The **y-intercept** is the value of f(n)f(n)f(n) when n=0n = 0n=0.However, since our sequence starts at n=1n = 1n=1, there's no y-intercept.

STEP 12

The **x-intercept** is the value of nnn when f(n)=0f(n) = 0f(n)=0.So, we set 0=42−7(n−1)0 = 42 - 7(n-1)0=42−7(n−1) and solve for nnn.

STEP 13

0=42−7(n−1)0 = 42 - 7(n-1)0=42−7(n−1).Subtract 42 from both sides: −42=−7(n−1)-42 = -7(n-1)−42=−7(n−1).

STEP 14

Divide both sides by -7: 6=n−16 = n - 16=n−1.

STEP 15

Add 1 to both sides: n=7n = 7n=7.So, the **x-intercept** is **7**.

STEP 16

Notice that as nnn increases, f(n)f(n)f(n) decreases.This means the sequence is **decreasing**.

STEP 17

Since the sequence is defined only for integer values of nnn, it's a **discrete** sequence.We can't have a term between, say, the first and second term.

STEP 18

First five terms: 42, 35, 28, 21, 14 Domain: {1, 2, 3, 4} Range: {42, 35, 28, 21} x-intercept: 7 y-intercept: None Intervals of increase: None Intervals of decrease: All Discrete/Continuous: Discrete

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1. Calculate the first five terms
2. Determine the domain and range
3. Find the x and y intercepts
4. Identify intervals of increase or decrease
5. Determine if the sequence is discrete or continuous

Let's calculate the first five terms!
We'll plug in $n = 1, 2, 3, 4, 5$ into our formula $f(n) = 42 - 7(n-1)$ .

For $n = 1$ , $f(1) = 42 - 7(1-1) = 42 - 7 \cdot 0 = 42 - 0 = 42$ .
Our first term is 42!

For $n = 2$ , $f(2) = 42 - 7(2-1) = 42 - 7 \cdot 1 = 42 - 7 = 35$ .
Our second term is 35!

For $n = 3$ , $f(3) = 42 - 7(3-1) = 42 - 7 \cdot 2 = 42 - 14 = 28$ . 28 is our third term!

For $n = 4$ , $f(4) = 42 - 7(4-1) = 42 - 7 \cdot 3 = 42 - 21 = 21$ .
The fourth term is 21!

For $n = 5$ , $f(5) = 42 - 7(5-1) = 42 - 7 \cdot 4 = 42 - 28 = 14$ .
And finally, our fifth term is 14!

The domain is the set of inputs, which are the first four $n$ values: $\{1, 2, 3, 4\}$ .

The range is the set of outputs, which are the first four terms we calculated: $\{42, 35, 28, 21\}$ .

The y-intercept is the value of $f(n)$ when $n = 0$ .
However, since our sequence starts at $n = 1$ , there's no y-intercept.

The x-intercept is the value of $n$ when $f(n) = 0$ .
So, we set $0 = 42 - 7(n-1)$ and solve for $n$ .

$0 = 42 - 7(n-1)$ .
Subtract 42 from both sides: $-42 = -7(n-1)$ .

Divide both sides by -7: $6 = n - 1$ .

Add 1 to both sides: $n = 7$ .
So, the x-intercept is 7.

Notice that as $n$ increases, $f(n)$ decreases.
This means the sequence is decreasing.

Since the sequence is defined only for integer values of $n$ , it's a discrete sequence.
We can't have a term between, say, the first and second term.