Math / Algebra

QuestionWhere is the graph of $f(x)=4\lfloor x-3\rfloor+2$ discontinuous? all real numbers all integers only at multiples of 3 only at multiples of 4

Studdy Solution

STEP 1

What is this asking? Where does the graph of the function $f(x) = 4\lfloor x-3\rfloor + 2$ suddenly jump or break? Watch out! The floor function can be tricky!
It's easy to get confused about where the jumps happen.

STEP 2

1. Understand the floor function
2. Analyze the given function
3. Determine points of discontinuity

STEP 3

Let's break down what the floor function, denoted by $\lfloor x \rfloor$ , actually does.
It takes any number $x$ and rounds it down to the nearest integer.
So, $\lfloor 3.2 \rfloor = 3$ , $\lfloor 5 \rfloor = 5$ , and $\lfloor -1.7 \rfloor = -2$ .

STEP 4

The key thing about the floor function is that it "jumps" at every integer value.
Think about it: as $x$ goes from $2.999$ to $3$ , $\lfloor x \rfloor$ jumps from $2$ to $3$ !
That's a discontinuity!

STEP 5

Our function is $f(x) = 4\lfloor x-3\rfloor + 2$ .
Let's see what's happening inside those floor function brackets: $x-3$ .
This means we shift the graph of $\lfloor x \rfloor$ three units to the right.

STEP 6

Now, what about that $4$ being multiplied outside?
This stretches the graph vertically by a factor of four.

STEP 7

Finally, adding $2$ shifts the entire graph two units up.

STEP 8

We know that the basic floor function $\lfloor x \rfloor$ is discontinuous at every integer.
Since we're just shifting and stretching, the discontinuities will still happen when the inside of the floor function, $x-3$ , is an integer.

STEP 9

So, when is $x-3$ an integer?
Well, when $x$ is an integer!
If $x$ is an integer, then $x-3$ will also be an integer.
For example, if $x=5$ , then $x-3=2$ , which is an integer.
If $x=1.5$ , then $x-3=-1.5$ , which is not an integer.

STEP 10

Therefore, the function $f(x) = 4\lfloor x-3\rfloor + 2$ is discontinuous at all integer values of $x$ .

STEP 11

The graph of $f(x) = 4\lfloor x-3\rfloor + 2$ is discontinuous at all integers.

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QuestionWhere is the graph of $f(x)=4\lfloor x-3\rfloor+2$ discontinuous? all real numbers all integers only at multiples of 3 only at multiples of 4

Studdy Solution

STEP 1

What is this asking? Where does the graph of the function $f(x) = 4\lfloor x-3\rfloor + 2$ suddenly jump or break? Watch out! The floor function can be tricky!
It's easy to get confused about where the jumps happen.

STEP 2

1. Understand the floor function
2. Analyze the given function
3. Determine points of discontinuity

STEP 3

Let's break down what the floor function, denoted by $\lfloor x \rfloor$ , actually does.
It takes any number $x$ and rounds it down to the nearest integer.
So, $\lfloor 3.2 \rfloor = 3$ , $\lfloor 5 \rfloor = 5$ , and $\lfloor -1.7 \rfloor = -2$ .

STEP 4

The key thing about the floor function is that it "jumps" at every integer value.
Think about it: as $x$ goes from $2.999$ to $3$ , $\lfloor x \rfloor$ jumps from $2$ to $3$ !
That's a discontinuity!

STEP 5

Our function is $f(x) = 4\lfloor x-3\rfloor + 2$ .
Let's see what's happening inside those floor function brackets: $x-3$ .
This means we shift the graph of $\lfloor x \rfloor$ three units to the right.

STEP 6

Now, what about that $4$ being multiplied outside?
This stretches the graph vertically by a factor of four.

STEP 7

Finally, adding $2$ shifts the entire graph two units up.

STEP 8

We know that the basic floor function $\lfloor x \rfloor$ is discontinuous at every integer.
Since we're just shifting and stretching, the discontinuities will still happen when the inside of the floor function, $x-3$ , is an integer.

STEP 9

So, when is $x-3$ an integer?
Well, when $x$ is an integer!
If $x$ is an integer, then $x-3$ will also be an integer.
For example, if $x=5$ , then $x-3=2$ , which is an integer.
If $x=1.5$ , then $x-3=-1.5$ , which is not an integer.

STEP 10

Therefore, the function $f(x) = 4\lfloor x-3\rfloor + 2$ is discontinuous at all integer values of $x$ .

STEP 11

The graph of $f(x) = 4\lfloor x-3\rfloor + 2$ is discontinuous at all integers.

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QuestionWhere is the graph of f(x)=4⌊x−3⌋+2f(x)=4\lfloor x-3\rfloor+2f(x)=4⌊x−3⌋+2 discontinuous? all real numbers all integers only at multiples of 3 only at multiples of 4

Studdy Solution

STEP 1

What is this asking? Where does the graph of the function f(x)=4⌊x−3⌋+2f(x) = 4\lfloor x-3\rfloor + 2f(x)=4⌊x−3⌋+2 suddenly jump or break? Watch out! The floor function can be tricky!It's easy to get confused about where the jumps happen.

STEP 2

1. Understand the floor function2. Analyze the given function3. Determine points of discontinuity

STEP 3

STEP 4

The key thing about the floor function is that it "jumps" at every integer value.Think about it: as xxx goes from 2.9992.9992.999 to 333, ⌊x⌋\lfloor x \rfloor⌊x⌋ jumps from 222 to 333!That's a discontinuity!

STEP 5

Our function is f(x)=4⌊x−3⌋+2f(x) = 4\lfloor x-3\rfloor + 2f(x)=4⌊x−3⌋+2.Let's see what's happening inside those floor function brackets: x−3x-3x−3.This means we **shift** the graph of ⌊x⌋\lfloor x \rfloor⌊x⌋ **three units to the right**.

STEP 6

Now, what about that 444 being multiplied outside?This **stretches** the graph vertically by a factor of **four**.

STEP 7

Finally, adding 222 **shifts** the entire graph **two units up**.

STEP 8

We know that the basic floor function ⌊x⌋\lfloor x \rfloor⌊x⌋ is discontinuous at every integer.Since we're just shifting and stretching, the discontinuities will still happen when the *inside* of the floor function, x−3x-3x−3, is an integer.

STEP 9

So, when is x−3x-3x−3 an integer?Well, when xxx is an integer!If xxx is an integer, then x−3x-3x−3 will also be an integer.For example, if x=5x=5x=5, then x−3=2x-3=2x−3=2, which is an integer.If x=1.5x=1.5x=1.5, then x−3=−1.5x-3=-1.5x−3=−1.5, which is not an integer.

STEP 10

Therefore, the function f(x)=4⌊x−3⌋+2f(x) = 4\lfloor x-3\rfloor + 2f(x)=4⌊x−3⌋+2 is discontinuous at all integer values of xxx.

STEP 11

The graph of f(x)=4⌊x−3⌋+2f(x) = 4\lfloor x-3\rfloor + 2f(x)=4⌊x−3⌋+2 is discontinuous at all integers.

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionWhere is the graph of $f(x)=4\lfloor x-3\rfloor+2$ discontinuous? all real numbers all integers only at multiples of 3 only at multiples of 4

What is this asking? Where does the graph of the function $f(x) = 4\lfloor x-3\rfloor + 2$ suddenly jump or break? Watch out! The floor function can be tricky!
It's easy to get confused about where the jumps happen.

1. Understand the floor function
2. Analyze the given function
3. Determine points of discontinuity

The key thing about the floor function is that it "jumps" at every integer value.
Think about it: as $x$ goes from $2.999$ to $3$ , $\lfloor x \rfloor$ jumps from $2$ to $3$ !
That's a discontinuity!

Our function is $f(x) = 4\lfloor x-3\rfloor + 2$ .
Let's see what's happening inside those floor function brackets: $x-3$ .
This means we shift the graph of $\lfloor x \rfloor$ three units to the right.

Now, what about that $4$ being multiplied outside?
This stretches the graph vertically by a factor of four.

Finally, adding $2$ shifts the entire graph two units up.

We know that the basic floor function $\lfloor x \rfloor$ is discontinuous at every integer.
Since we're just shifting and stretching, the discontinuities will still happen when the inside of the floor function, $x-3$ , is an integer.

So, when is $x-3$ an integer?
Well, when $x$ is an integer!
If $x$ is an integer, then $x-3$ will also be an integer.
For example, if $x=5$ , then $x-3=2$ , which is an integer.
If $x=1.5$ , then $x-3=-1.5$ , which is not an integer.

Therefore, the function $f(x) = 4\lfloor x-3\rfloor + 2$ is discontinuous at all integer values of $x$ .

The graph of $f(x) = 4\lfloor x-3\rfloor + 2$ is discontinuous at all integers.