# Thread: Find f(-1), f(1), f(3)

1. ## Find f(-1), f(1), f(3)

Here is another piecewise defined function.

Find f(-1), f(1), f(3)

.........{x + 3.....-2<or=tox<1
f(x)=..{3...........x = 1
.........{-x + 3....x > 1

I need someone to explain this question piece by piece.
I really never really understood this type of function.

Thanks

2. To understand a piece wise function is simple. Whenever you are given a value of x, to know f(x), see which of the given range x belongs to and then substitute for x in that equation...

Here say I want f(1), then observe that the definition says f(x) = 3 when x = 1, which means f(1) = 3.

To know f(-1), observe that definition says whenever x is between -2 and 1, f(x) is x+3. So f(-1) = -1+3 = 2

Now can you try f(3)?

3. ## but....

Originally Posted by Isomorphism
To understand a piece wise function is simple. Whenever you are given a value of x, to know f(x), see which of the given range x belongs to and then substitute for x in that equation...

Here say I want f(1), then observe that the definition says f(x) = 3 when x = 1, which means f(1) = 3.

To know f(-1), observe that definition says whenever x is between -2 and 1, f(x) is x+3. So f(-1) = -1+3 = 2

Now can you try f(3)?
I know that the only number that pertain f(x) = 3, has to be x = 1. So, when x = 1, f(x) = 3 or y = 3.

However, you said:

"To know f(-1), observe that definition says whenever x is between -2 and 1, f(x) is x+3. So f(-1) = -1+3 = 2"

The retriction given is that x lies between -2 and 1. If I replace x with -1 for the function f(x) = x + 3, then how can x be lies between -2 and 1?

Look:

-1 + 3 = 2

Is 2 less than 1?

Here is the restriction:

-2 <or= to x < 1

If I place 2 for x in the restriction, then the inequality does not make sense.

I thought that we can replace x with -1 for the function
f(x) = -x + 3 where the restriction is x > 1.

f(-1) = -(-1) + 3

f(-1) = 1 + 3

f(-1) = 4

I can see that 4 > 1 is true.

See my point?

Thanks

4. Originally Posted by magentarita
Look:

-1 + 3 = 2

Is 2 less than 1?

Here is the restriction:

-2 <or= to x < 1

If I place 2 for x in the restriction, then the inequality does not make sense.
The question says f(x) = x+3 when x is between -2 and 1.

Your computation -1 + 3 is f(-1), right? The definition does not say f(-1) is between -2 and 1.

Understand?

5. ## got it now....

Originally Posted by Isomorphism
The question says f(x) = x+3 when x is between -2 and 1.

Your computation -1 + 3 is f(-1), right? The definition does not say f(-1) is between -2 and 1.

Understand?
I see what you mean.

We are applying what lies between the parentheses for f(x) to the proper restriction.

In that case, it makes sense that f(-1) be evaluated for x + 3 because
-1 lies between -2 and 1.

I see.

Thanks