How Do I Graph a Piecewise Function?

[skyza]

Can anyone help me graph this function? I don't think we took notes on this in class.



f(x)={

7x+7
if x ≤ 0

x
if 0 < x ≤ 2

4
if x ≥ 2





Thank you

 

[spaghetti3451]

Draw the x and y axes, first of all.

For x ≤ 0, find f(-1) and f(-2) and draw a line that passes through (-1, f(-1)) and (-2, f(-2)) s.t. the line extends throughout x ≤ 0.

For 0 < x ≤ 2, find f(1) and f(2) and draw a line that passes through (1, f(1)) and (2, f(2)) s.t. the line extends throughout 0 < x ≤ 2.

For x > 2, f(x) = 4. In other words, no matter what the value of x, y is always 4. Therefore, draw a horizontal line s.t. y is 4.

 

[RoshanBBQ]

A function is a correspondence between the input to the function, which lies in its domain, and the value the function returns for that given input, which lies in its range. We normally think of x as an input to a function and y as its output. The functional relationship is written as y = f(x). That is, the output is the value returned by the function whose input is x, some value that lies within f's domain.

Your function is piecewise, so it has a different correspondence between input and output for certain parts of the domain. If x, the input, is less than or equal to 0, your functional relationship between y and x is 7x+7. So for this region on your graph, which is a visual representation of a functional relation, you would plot y = 7x + 7.

 

[skyza]

Draw the x and y axes, first of all.

For x ≤ 0, find f(-1) and f(-2) and draw a line that passes through (-1, f(-1)) and (-2, f(-2)) s.t. the line extends throughout x ≤ 0.

For 0 < x ≤ 2, find f(1) and f(2) and draw a line that passes through (1, f(1)) and (2, f(2)) s.t. the line extends throughout 0 < x ≤ 2.

For x > 2, f(x) = 4. In other words, no matter what the value of x, y is always 4. Therefore, draw a horizontal line s.t. y is 4.

Thank you!

I was close. Are there any rules about filling in the circle where you mark your points?
like ° or •

 

[RoshanBBQ]

In a piecewise definition of a function, there are points where the functional relation between input and output changes. In your problem, one such point is at x = 0 where the functional relation transitions from y = 7x+7 to y=x. We see that evaluating both of these relations at zero yields different results. So what is the value of y = f(0)? Is it:
y = 7*0 + 7 = 7
or
y = 0 = 0

We see that y = 7x+7 for when x is less than or equal to 0. So the function takes on 7 at x = 0 since 0 is less than or equal to 0. We denote this behavior during a transition between two functional relations in a piecewise function with a filled circle at x=0, y = 7. An unfilled circle will also be at x = 0, y = 0. That is because y = x only when x is greater than 0. So the function does not actually take the value y = 0 at x = 0 since 0 is not greater than zero.

On a different note, if the two values for the two correspondences equal, you needn't any circle. We call this a continuous function, and you would draw past that point without picking up your pencil. A trivial case would be if we defined
<br /> y =f(x) =<br /> \begin{cases}<br /> x, &amp; \text{if }x\geq0 \\<br /> x, &amp; \text{if }x&lt;0<br /> \end{cases}<br />
This equation would be a straight line through the origin without any filled or unfilled circles at x = 0 since both definitions at x = 0 return 0.

 

[skyza]

Thanks for the help.
I have one other quick question.

Would the requested value for this problem be 14?

f(1) for

f(x) = {

7x+7
if x ≤ 0

3-2x^2
if 0 < x ≤ 2

4
if x ≥ 2

 

[RoshanBBQ]

Thanks for the help.
I have one other quick question.

Would the requested value for this problem be 14?

f(1) for

f(x) = {

7x+7
if x ≤ 0

3-2x^2
if 0 < x ≤ 2

4
if x ≥ 2

Is 1 less than or equal to 0? That is a yes or no question -- not an either this or that question.

 

[skyza]

Is 1 less than or equal to 0? That is a yes or no question -- not an either this or that question.

Yes. So the answer is 1, right?

 

[RoshanBBQ]

Yes. So the answer is 1, right?

Is it your opinion that 1 is less than 0 or equal to 0? This one is an either or question.

 

[skyza]

Is it your opinion that 1 is less than 0 or equal to 0? This one is an either or question.

Sorry, it's late. I read that wrong. 1 is greater than 0.

 

[RoshanBBQ]

Sorry, it's late. I read that wrong. 1 is greater than 0.

It's not a problem. I am trying to work with you to think through the problems. If 1 is greater than 0, it is not less than or equal to 0. Which functional relationship should you use then?

 

[skyza]

It's not a problem. I am trying to work with you to think through the problems. If 1 is greater than 0, it is not less than or equal to 0. Which functional relationship should you use then?

0 < x ≤ 2

 

[RoshanBBQ]

0 < x ≤ 2

That isn't the functional relationship between y and x. We wouldn't say y = f(x) = 0 < x ≤ 2. That is the condition that brings about from which functional relationship we choose. So you did choose the right condition. The functional relationship is right beside it: 3-2x^2.

That is, between 0 < x ≤ 2, we say y = 3-2x^2. This equation guides us in mapping input, x, to output, y. For any x in that range, we know the output by plugging it straight into that equation.

 

[skyza]

That isn't the functional relationship between y and x. We wouldn't say y = f(x) = 0 < x ≤ 2. That is the condition that brings about from which functional relationship we choose. So you did choose the right condition. The functional relationship is right beside it: 3-2x^2.

That is, between 0 < x ≤ 2, we say y = 3-2x^2.

So how would I find out the answer?

 

[RoshanBBQ]

So how would I find out the answer?

You want to know what the function, f(x) outputs at x = 1. For 0 < x ≤ 2, we say the function outputs 3-2x^2. You have confirmed that 1 is between 0 and 2. Do you know how to use this functional relationship to output the answer? Hint: x is just a placeholder in 3-2x^2. What should you put there?

 

[skyza]

you want to know what the function, f(x) outputs at x = 1. For 0 < x ≤ 2, we say the function outputs 3-2x^2. You have confirmed that 1 is between 0 and 2. Do you know how to use this functional relationship to output the answer? Hint: x is just a placeholder in 3-2x^2. What should you put there?

3-2(1)^2 = -1

 

[RoshanBBQ]

3-2(1)^2 = -1

You have the right idea, but you messed up in the maneuver. Think about what a squaring of a number does: It is a shorthand way to say "Let's do some multiplication." When you have
1^2
you are saying "Let's multiply two 1s together. For example, if you had
3^3
You would be saying, "Let's multiply three 3s together. So
3^3 = (3)(3)(3) = (9)(3) = 27

Another example would be if I had
2^4
That is saying "Let's multiply four 2s together.
(2)(2)(2)(2) = (4)(2)(2)=(8)(2) = 16

What is 1 multiplied together two times?

 

[skyza]

You have the right idea, but you messed up in the maneuver. Think about what a squaring of a number does: It is a shorthand way to say "Let's do some multiplication." When you have
1^2
you are saying "Let's multiply two 1s together. For example, if you had
3^3
You would be saying, "Let's multiply three 3s together. So
3^3 = (3)(3)(3) = (9)(3) = 27

Another example would be if I had
2^4
That is saying "Let's multiply four 2s together.
(2)(2)(2)(2) = (4)(2)(2)=(8)(2) = 16

What is 1 multiplied together two times?

it's 1

 

[RoshanBBQ]

it's 1

So what is 3 - (2)(1)?

 

[skyza]

So what is 3 - (2)(1)?

1. Is 1 the answer?

 

[RoshanBBQ]

1. Is 1 the answer?

Yes.

 

[skyza]

Yes.

Thanks for the help.

I've finished all of my homework except ONE question.

What is the range of this question:
f(x) = √(x-1) +1

I know it's not (-∞, ∞)

Is it [1, ∞)?

 

[RoshanBBQ]

Thanks for the help.

I've finished all of my homework except ONE question.

What is the range of this question:
f(x) = √(x-1) +1

I know it's not (-∞, ∞)

Is it [1, ∞)?

Yes.

 

[skyza]

Yes.

I really appreciate all the help. I'm going to school to become a mechanical engineer, so I'm sure I'll be back with more questions.

 

[RoshanBBQ]

I really appreciate all the help. I'm going to school to become a mechanical engineer, so I'm sure I'll be back with more questions.

No problem. It's good that you put thought into your work.