Problem Analysis:
The image shows a page from a precalculus textbook focusing on
average rate of change and
concavity. The task involves analyzing the function \( h(x) = -x^3 \) over specific intervals and determining its behavior in terms of concavity, average rate of change, and inflection points.
#### Key Concepts:
1.
Concavity:
- A function is
positively concave (concave up) if its slope increases as \( x \) increases.
- A function is
negatively concave (concave down) if its slope decreases as \( x \) increases.
- An
inflection point occurs where the concavity changes.
2.
Average Rate of Change:
\[
\text{Average Rate of Change} = \frac{h(b) - h(a)}{b - a}
\]
This measures how much the function changes per unit change in \( x \) over the interval \([a, b]\).
---
Task Breakdown:
We are tasked with analyzing the function \( h(x) = -x^3 \) over specific intervals and answering related questions.
#### Part (i): Behavior of \( h(x) \) as \( x \) increases from \(-4\) to \(-2.5\)
-
Step 1: Compute \( h(-4) \) and \( h(-2.5) \).
\[
h(-4) = -(-4)^3 = -(-64) = 64
\]
\[
h(-2.5) = -(-2.5)^3 = -(-15.625) = 15.625
\]
-
Step 2: Determine the change in \( h(x) \).
\[
h(x) \text{ changes from } 64 \text{ to } 15.625.
\]
#### Part (ii): Behavior of \( h(x) \) as \( x \) increases from \(-2.5\) to \(-1\)
-
Step 1: Compute \( h(-2.5) \) and \( h(-1) \).
\[
h(-2.5) = 15.625 \quad (\text{already computed})
\]
\[
h(-1) = -(-1)^3 = -(-1) = 1
\]
-
Step 2: Determine the change in \( h(x) \).
\[
h(x) \text{ changes from } 15.625 \text{ to } 1.
\]
#### Part (iii): Average Rate of Change of \( h(x) \) from \( x = -4 \) to \( x = -2.5 \)
-
Step 1: Use the formula for average rate of change.
\[
\text{Average Rate of Change} = \frac{h(-2.5) - h(-4)}{-2.5 - (-4)}
\]
\[
= \frac{15.625 - 64}{-2.5 + 4} = \frac{-48.375}{1.5} = -32.25
\]
-
Step 2: Interpret the result.
The average rate of change is \(-32.25\), indicating that \( h(x) \) decreases by 32.25 units per unit increase in \( x \) over this interval.
#### Part (iv): Average Rate of Change of \( h(x) \) from \( x = -2.5 \) to \( x = -1 \)
-
Step 1: Use the formula for average rate of change.
\[
\text{Average Rate of Change} = \frac{h(-1) - h(-2.5)}{-1 - (-2.5)}
\]
\[
= \frac{1 - 15.625}{-1 + 2.5} = \frac{-14.625}{1.5} = -9.75
\]
-
Step 2: Interpret the result.
The average rate of change is \(-9.75\), indicating that \( h(x) \) decreases by 9.75 units per unit increase in \( x \) over this interval.
#### Part (v): Compare the Average Rates of Change
- From part (iii), the average rate of change from \( x = -4 \) to \( x = -2.5 \) is \(-32.25\).
- From part (iv), the average rate of change from \( x = -2.5 \) to \( x = -1 \) is \(-9.75\).
- Since \(-32.25 < -9.75\), the average rate of change is
greater (in absolute value) from \( x = -4 \) to \( x = -2.5 \).
---
Final Answer:
\[
\boxed{\text{T}}
\]
Parent Tip: Review the logic above to help your child master the concept of precalculus workbook with answers.
