Question

chapter 4
polynomial functions

1. use finite differences to determine the degree of

the polynomial function that fits the data. then use
technology to find the polynomial function.

x	f(x)
2	-24
3	-27
4	-8
5	45
6	144

Explanation:

Step 1: Find First Differences

First, we calculate the first differences (differences between consecutive \( f(x) \) values):

• Between \( x = 1 \) and \( x = 2 \): \( -24 - (-11) = -13 \)
• Between \( x = 2 \) and \( x = 3 \): \( -27 - (-24) = -3 \)
• Between \( x = 3 \) and \( x = 4 \): \( -8 - (-27) = 19 \)
• Between \( x = 4 \) and \( x = 5 \): \( 45 - (-8) = 53 \)
• Between \( x = 5 \) and \( x = 6 \): \( 144 - 45 = 99 \)

First differences: \( -13, -3, 19, 53, 99 \)

Step 2: Find Second Differences

Now, calculate the second differences (differences between consecutive first differences):

• Between \( -13 \) and \( -3 \): \( -3 - (-13) = 10 \)
• Between \( -3 \) and \( 19 \): \( 19 - (-3) = 22 \)
• Between \( 19 \) and \( 53 \): \( 53 - 19 = 34 \)
• Between \( 53 \) and \( 99 \): \( 99 - 53 = 46 \)

Second differences: \( 10, 22, 34, 46 \)

Step 3: Find Third Differences

Calculate the third differences (differences between consecutive second differences):

• Between \( 10 \) and \( 22 \): \( 22 - 10 = 12 \)
• Between \( 22 \) and \( 34 \): \( 34 - 22 = 12 \)
• Between \( 34 \) and \( 46 \): \( 46 - 34 = 12 \)

Third differences: \( 12, 12, 12 \)

Since the third differences are constant (all equal to 12), the degree of the polynomial is equal to the level of differences that became constant. Here, the third differences are constant, so the degree of the polynomial is 3.

To find the polynomial function, we can use the general form of a cubic polynomial \( f(x) = ax^3 + bx^2 + cx + d \). We can set up a system of equations using the given \( x \) and \( f(x) \) values.

Using \( x = 1 \): \( a(1)^3 + b(1)^2 + c(1) + d = -11 \) ⇒ \( a + b + c + d = -11 \)

Using \( x = 2 \): \( a(2)^3 + b(2)^2 + c(2) + d = -24 \) ⇒ \( 8a + 4b + 2c + d = -24 \)

Using \( x = 3 \): \( a(3)^3 + b(3)^2 + c(3) + d = -27 \) ⇒ \( 27a + 9b + 3c + d = -27 \)

Using \( x = 4 \): \( a(4)^3 + b(4)^2 + c(4) + d = -8 \) ⇒ \( 64a + 16b + 4c + d = -8 \)

Now, we can solve this system of equations.

Subtract the first equation from the second:
\( (8a + 4b + 2c + d) - (a + b + c + d) = -24 - (-11) \)
\( 7a + 3b + c = -13 \) --- Equation (1)

Subtract the second equation from the third:
\( (27a + 9b + 3c + d) - (8a + 4b + 2c + d) = -27 - (-24) \)
\( 19a + 5b + c = -3 \) --- Equation (2)

Subtract the third equation from the fourth:
\( (64a + 16b + 4c + d) - (27a + 9b + 3c + d) = -8 - (-27) \)
\( 37a + 7b + c = 19 \) --- Equation (3)

Now, subtract Equation (1) from Equation (2):
\( (19a + 5b + c) - (7a + 3b + c) = -3 - (-13) \)
\( 12a + 2b = 10 \) ⇒ \( 6a + b = 5 \) --- Equation (4)

Subtract Equation (2) from Equation (3):
\( (37a + 7b + c) - (19a + 5b + c) = 19 - (-3) \)
\( 18a + 2b = 22 \) ⇒ \( 9a + b = 11 \) --- Equation (5)

Subtract Equation (4) from Equation (5):
\( (9a + b) - (6a + b) = 11 - 5 \)
\( 3a = 6 \) ⇒ \( a = 2 \)

Substitute \( a = 2 \) into Equation (4):
\( 6(2) + b = 5 \) ⇒ \( 12 + b = 5 \) ⇒ \( b = -7 \)

Substitute \( a = 2 \) and \( b = -7 \) into Equation (1):
\( 7(2) + 3(-7) + c = -13 \)
\( 14 - 21 + c = -13 \)
\( -7 + c = -13 \) ⇒ \( c = -6 \)

Substitute \( a = 2 \), \( b = -7 \), and \( c = -6 \) into the first equation \( a + b + c + d = -11 \):
\( 2 - 7 - 6 + d = -11 \)
\( -11 + d = -11 \) ⇒ \( d = 0 \)

So, the polynomial function is \( f(x) = 2x^3 - 7x^2 - 6x \)

Answer:

The degree of the polynomial is 3, and the polynomial function is \( f(x) = 2x^3 - 7x^2 - 6x \)