Question

what is the rate of change of the average rates of change for each function over consecutive equal-length intervals?

1. ( w(x) = 5 - 3x )
2. ( f(x) = -3x^2 + 6x - 1 )

hw1.3a # 9,10,11
( \begin{array}{c|c} w(x) & 5 - 3x \\ hline -2 & -2.33 -1.99 \\ -1 & 3, -1 \\ 0 & 0, 1.666 \\ 1 & 1.33, 1.01 \\ 2 & 1, 2 \\ end{array} )
( \begin{array}{c|c} f(x) & -3x^2 + 6x - 1 \\ hline -2 & -25 \\ -1 & -10 \\ 0 & -1 \\ 1 & 2 \\ 2 & 4 \\ end{array} )

Explanation:

Problem 11: \( w(x) = 5 - 3x \)

Step 1: Recall the formula for average rate of change

The average rate of change of a function \( w(x) \) over the interval \([a, a + h]\) is given by \(\frac{w(a + h) - w(a)}{h}\).

Step 2: Calculate \( w(a + h) \) and \( w(a) \)

Given \( w(x) = 5 - 3x \), we have:
\( w(a + h) = 5 - 3(a + h) = 5 - 3a - 3h \)
\( w(a) = 5 - 3a \)

Step 3: Compute the average rate of change

The average rate of change over \([a, a + h]\) is:
\[
\frac{w(a + h) - w(a)}{h} = \frac{(5 - 3a - 3h) - (5 - 3a)}{h} = \frac{-3h}{h} = -3
\]

Step 4: Find the rate of change of the average rates of change

Since the average rate of change is a constant (\(-3\)) for all intervals, the rate of change of the average rates of change is \( 0 \) (because the average rate of change does not change).

Step 1: Recall the formula for average rate of change

The average rate of change of a function \( f(x) \) over the interval \([a, a + h]\) is \(\frac{f(a + h) - f(a)}{h}\).

Step 2: Calculate \( f(a + h) \) and \( f(a) \)

Given \( f(x) = -3x^2 + 6x - 1 \), we have:
\( f(a + h) = -3(a + h)^2 + 6(a + h) - 1 = -3(a^2 + 2ah + h^2) + 6a + 6h - 1 = -3a^2 - 6ah - 3h^2 + 6a + 6h - 1 \)
\( f(a) = -3a^2 + 6a - 1 \)

Step 3: Compute the average rate of change

The average rate of change over \([a, a + h]\) is:
\[

$$\begin{align*} \frac{f(a + h) - f(a)}{h} &= \frac{(-3a^2 - 6ah - 3h^2 + 6a + 6h - 1) - (-3a^2 + 6a - 1)}{h} \\ &= \frac{-6ah - 3h^2 + 6h}{h} \\ &= -6a - 3h + 6 \end{align*}$$

\]

Step 4: Find the rate of change of the average rates of change

Now, we find the average rate of change of the average rate of change. Let's take two consecutive intervals, say with \( a_1 \) and \( a_2 = a_1 + h \).

First, the average rate of change at \( a_1 \) is \( -6a_1 - 3h + 6 \).

The average rate of change at \( a_2 = a_1 + h \) is \( -6(a_1 + h) - 3h + 6 = -6a_1 - 6h - 3h + 6 = -6a_1 - 9h + 6 \).

The rate of change of the average rates of change over the interval \([a_1, a_1 + h]\) is:
\[
\frac{(-6a_1 - 9h + 6) - (-6a_1 - 3h + 6)}{h} = \frac{-6h}{h} = -6
\]

(Alternatively, since the original function is quadratic, the rate of change of the average rate of change is twice the coefficient of \( x^2 \) times the interval length. For \( f(x) = ax^2 + bx + c \), the second difference (rate of change of average rate of change) over equal intervals of length \( h \) is \( 2ah^2 \). Here, \( a = -3 \), \( h = 1 \) (assuming unit intervals), so \( 2(-3)(1)^2 = -6 \), which matches.)

Answer:

\( 0 \)

Problem 12: \( f(x) = -3x^2 + 6x - 1 \)