Problem 47 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(f^{\prime}(x)=x^{3}-5 x^{2}+2\) on \(I=[-2,4]\). Where on \(I\) is \(f\) increasing?

Step-by-Step Solution

Answer

Function \(f(x)\) is increasing where \(x^{3} - 5x^{2} + 2 > 0\) on \([-2, 4]\).

Step 1: Understand when a function is increasing

A function is increasing on an interval where its derivative is greater than zero.

Step 2: Set derivative greater than zero

Given the derivative \(f^{\prime}(x)=x^{3}-5x^{2}+2\), we want to find the intervals where \(x^{3} - 5x^{2} + 2 > 0\).

Step 3: Solve the inequality

We solve the inequality \(x^{3} - 5x^{2} + 2 > 0\) by finding critical points where \(x^{3} - 5x^{2} + 2 = 0\) and testing the sign of the inequality between these points.

Step 4: Find critical points

Solve the equation \(x^{3} - 5x^{2} + 2 = 0\) to identify critical points. In this polynomial, use techniques like factoring or the Rational Root Theorem if applicable.

Step 5: Test intervals

Use the critical points to divide the interval \([-2,4]\) into subintervals and test a value from each to determine where \(f^{\prime}(x) > 0\).

Step 6: Identify where function is increasing

After testing, determine which subintervals have \(x^{3} - 5x^{2} + 2 > 0\). These subintervals are where the original function \(f(x)\) is increasing.

Key Concepts

Derivative

In calculus, a derivative represents how a function changes as its input changes. Think of it as the function's rate of change or slope at any given point. When we talk about the derivative of a function, such as \( f'(x) = x^3 - 5x^2 + 2 \), we imply that this expression tells us how the original function \( f(x) \) behaves across different values of \( x \).

The process of finding a derivative is called differentiation. This mathematical operation helps us understand important properties of functions, such as intervals where the function is increasing or decreasing. By analyzing \( f'(x) \), we can infer the local behavior of \( f(x) \) around different points.

Increasing Function

A function \( f(x) \) is said to be increasing on an interval where its derivative, \( f'(x) \), is greater than zero. This means the function's value is rising as you move along the interval. For instance, if we determine that \( f'(x) > 0 \) for specific ranges of \( x \), it means the function \( f(x) \) climbs upward across those ranges.

To find these intervals for our specific derivative \( x^3 - 5x^2 + 2 \), we:

• Determine where the derivative is positive.
• Solve the inequality \( x^3 - 5x^2 + 2 > 0 \) to find intervals of increase.

Using this method, such regions indicate where the function exhibits growth.

Critical Points

Critical points occur where the derivative equals zero or is undefined, impacting whether a function increases or decreases in a nearby neighborhood. They serve as useful markers, especially within the interval \( I = [-2, 4] \), guiding us in locating potential changes in the function's direction.

To find critical points for \( f'(x) = x^3 - 5x^2 + 2 \), solve the equation:

• \( x^3 - 5x^2 + 2 = 0 \)

These solutions partition the interval into segments, simplifying our task of determining the derivative's sign in each segment. Checking the sign of \( f'(x) \) in these partitions highlights sections where \( f(x) \) switches between increasing and decreasing behavior.

Polynomial Inequality

Solving polynomial inequalities like \( x^3 - 5x^2 + 2 > 0 \) involves identifying where a polynomial expression is positive or negative. These solutions are instrumental in understanding how the function behaves across different intervals.

Here's a simple approach:

• Identify the polynomial's critical points by solving \( x^3 - 5x^2 + 2 = 0 \).
• Divide the domain \( I = [-2, 4] \) based on these points.
• Pick test points from each subinterval to determine whether the expression is positive or negative in that segment.

Consequently, the intervals where the inequality holds true indicate the parts of the domain where \( f(x) \) is increasing, providing us with valuable insight into the function’s growth patterns.