8.3OtherPiecewiseFunctions
We have examined two piecewise functions so far. The absolute value linear function has two pieces
making the stark “V” shape, and the step functions had each piece looking like a step in a staircase. Now we will
examine generic piecewise defined functions. This means we define what function we are dealing over each piece
of the domain.
Overview of Piecewise Defined Functions
When we use a piecewise defined function, we normally will use the left curly bracket to contain the
different pieces of the function. We first list what the function is on that piece and then domain where that function
exists. For example, look at the following piecewise defined function:
2 4; f −2
− 4, ; −2 g f 2
Ê
1
− 1; / 2
2
In this case, the function is defined by three pieces. The first piece is a linear function. The function ()
looks and acts like a linear function on the domain f −2 (or to the left of = −2). However, exactly at = −2,
the function () begins to look and act like a quadratic. It continues this behavior until we get to the domain /
2 at which point it begins to look like a linear function again but a different one. Take a look at the graph from a TI84:
1
1
2
= 2 + 4
4
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It is also possible that the pieces don’t actually connect like they do in the above example. In the case where
they don’t connect, we would need some open and closed circles marking the end points depending on whether
the domain was an “or equal to” or not. Consider the following piecewise defined graph and its graph:
2 12; f 6
F Ë 1
6; / 6
3
Notice the open circle on the 2 12 portion of the function and the closed circle on the C 6 portion.
These are due to the inequalities associated with their domains.
Graphing Piecewise Defined Functions
So how exactly do we create these graphs? In essence, we have to graph each of the pieces separately but
only on the domain specified in the definitions. Let’s look at the above function F which had two pieces. In that
case, we would make two separate / charts but use inputs that are within the domain as follows:
F 2 12
Domain: f 6
F
10
8
F C 6
9
6
8
4
7
2
Since inputs must be
less than 6, this will
be an open circle.
6
0
Domain: / 6
F
6
4
3
5
0
6
3
7
6
8
Since inputs can be
equal to 6, this will
be a closed circle.
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In the case of our function , there were three pieces, so we would need three separate / charts
using inputs within the domain. The only difference is that for the function , each piece ends on the same
point as the next piece starts. That means we don’t have to worry about the open and closed circles since they
would overlap.
Open circle from left and
closed circle from right.
They overlap, so no circle
is necessary.
Let’s again use our function as follows:
Solving Piecewise Defined Equations
2 4; f −2
− 4, ; −2 g f 2
Ê
1
− 1; / 2
2
We can now talk about solving this as an equation if () = 5. To do so, we need to solve each piece of
the function separately as follows:
2 + 4 5
4 5
15
2 1
9
6
0.5
±3
2 4 4 5 4
1
4 4 5 4
√ √9
1151
2 ∗ 6 ∗2
12
We now have four separate answers (since the second answer is both the positive and the negative).
However, are each of these answers within their respective domains? For the first answer we got 0.5, but if
0.5 were our input, we would not be using that piece of the function. That piece of the function is only for when
f 2, and therefore 0.5 is not a valid solution. Similarly, we got an answer of ±3 for the quadratic
portion, but that’s only when is between 2 and 2. Therefore ±3 is not a valid solution. The last answer is
the only correct solution because 12 is within the domain of the piece of defined as 1.
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Lesson 8.3
Fill out an / chart and graph each of the following piecewise functions.
1. »
()
()
3. () =
()
()
C
5; / −3
−2 − 9; f −3
»
− 5; f 0
− 2; / 0
2. () = µ
()
()
4. () =
()
()
()
2; g 2
−( − 3) + 6; k 2
2 + 15; f −6
Ê− C + 1;−6 g g
− 4; k 6
6
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− − 4; f −2
− − 8 − 12; g −2
5. () = Å
−8;−2 f f 2
− + 8 − 12; / 2
6. () = Ê − 7;−2 g g 2
− 4; k 2
()
()
()
()
()
()
Solve the following equations.
2 5; g −5
7. 3 where »
− 4; k −5
( + 3) − 1; g 5
9. () = −2 where () = µ
− + 1; k 5
8. () = 4 where µ
4 8; / 0
− 2 − 4; f 0
5; f 1
10. () = 4 where Å−2 + 10; 1 g g 6
− 2; k 6
3 + 3; g −1
3 5; f −8
11. () = 0 where () = Å − 4 5;−1 f g 512. () = −7 where Ë 2 + 3;−8 g g 5
+ 4; k 5
−( − 3) + 9; k 5
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