Problem 29 - Essential Calculus Early Transcendentals

Question

(a) Find the intervals of increase or decrease. (b) Find the local maximum and minimum values. (c) Find the intervals of concavity and the inflection points. (d) Use the inforvation from parts (a)-(c) to sketch the graph. Check your work with a graphing device if you have one. \(h(x)=(x+1)^{5}-5 x-2\)

Step-by-Step Solution

Answer

(a) Increasing: \((-\infty, -2)\cup(0, \infty)\); Decreasing: \((-2, 0)\). (b) Local max at \(x=-2\); min at \(x=0\). (c) Concave down: \((-\infty, -1)\); concave up: \((-1, \infty)\); inflection: \(x=-1\).

Step 1: Find the Derivative

First, we need to find the first derivative of the function, as it will help us determine intervals of increase and decrease. The given function is: \( h(x) = (x+1)^5 - 5x - 2 \).Differentiate with respect to \( x \):\[ h'(x) = 5(x+1)^4 - 5. \]

Step 2: Solve for Critical Points

Set the first derivative equal to zero to find the critical points where the function might change from increasing to decreasing or vice versa:\[ 5(x+1)^4 - 5 = 0. \]Divide everything by 5:\[ (x+1)^4 - 1 = 0. \]Adding 1 to both sides:\[ (x+1)^4 = 1. \]Taking the fourth root:\[ x+1 = \pm 1. \]Then, solve for \( x \):\[ x = 0 \quad \text{or} \quad x = -2. \]

Step 3: Determine Intervals of Increase and Decrease

To determine where the function is increasing or decreasing, test the intervals defined by the critical points \( x = 0 \) and \( x = -2 \).1. For \( x < -2 \) (e.g., \( x = -3 \)), \( h'(-3) = 5(-2)^4 - 5 > 0 \); increasing.2. For \( -2 < x < 0 \) (e.g., \( x = -1 \)), \( h'(-1) = 5(0)^4 - 5 = -5 < 0 \); decreasing.3. For \( x > 0 \) (e.g., \( x = 1 \)), \( h'(1) = 5(2)^4 - 5 > 0 \); increasing.Thus, \( h(x) \) is increasing on \( (-\infty, -2) \cup (0, \infty) \) and decreasing on \( (-2, 0) \).

Step 4: Identify Local Extrema

Based on the intervals of increase and decrease:- At \( x = -2 \), the function changes from increasing to decreasing. Therefore, \( x = -2 \) is a local maximum.- At \( x = 0 \), the function changes from decreasing to increasing. Thus, \( x = 0 \) is a local minimum.

Step 5: Find the Second Derivative

Now, we need the second derivative to analyze concavity:Differentiate \( h'(x) = 5(x+1)^4 - 5 \):\[ h''(x) = 20(x+1)^3. \]

Step 6: Solve for Inflection Points

An inflection point occurs where the second derivative equals zero or is undefined, indicating a change in concavity:\[ 20(x+1)^3 = 0. \]Solving for \( x \):\[ (x+1)^3 = 0 \]\[ x = -1. \]Therefore, \( x = -1 \) is the inflection point.

Step 7: Determine Intervals of Concavity

Test the intervals divided by \( x = -1 \) using the second derivative:1. For \( x < -1 \) (e.g., \( x = -2 \)), \( h''(-2) = 20(-1)^3 = -20 < 0 \); concave down.2. For \( x > -1 \) (e.g., \( x = 0 \)), \( h''(0) = 20(1)^3 = 20 > 0 \); concave up.Thus, the function is concave down on \( (-\infty, -1) \) and concave up on \( (-1, \infty) \).

Step 8: Sketch the Graph Using Information

Based on previous steps, sketch the graph:- The function increases on \((-\infty, -2)\) and \((0, \infty)\). It decreases on \((-2, 0)\).- There is a local maximum at \(x = -2\) and a local minimum at \(x = 0\).- The inflection point at \(x = -1\) indicates a change from concave down to concave up.Utilize this information to draw the graph, noting the key points.

Key Concepts

Derivatives

In calculus, the derivative of a function is like a tool that tells us the rate at which the function's value changes as its input changes. Think of it as a measure of how steep or flat a curve is at a given point. It can help you understand whether a function is climbing upward, falling downward, or doing something else.

The derivative of a function, denoted as \( h'(x) \), is the main player when we're working on determining where a function increases or decreases. In our original example, the derivative of the function \( h(x) = (x+1)^5 - 5x - 2 \) was found to be \( h'(x) = 5(x+1)^4 - 5 \).

This derivative gives us plenty of information. By setting \( h'(x) = 0 \), we can find critical points that help us divide the function into different behavior regions. Therefore, understanding how to derive a function is a crucial skill in calculus graphing.

Critical Points

Critical points are specific values of \( x \) where the derivative of a function equals zero or is undefined. These points indicate potential changes in the behavior of the function.

In the exercise, after we found the derivative \( h'(x) = 5(x+1)^4 - 5 \), we solved for when this equals zero to find critical points. Solving \( (x+1)^4 - 1 = 0 \) led us to obtain \( x = 0 \) and \( x = -2 \) as critical points.

Critical points are essential because they may mark where a function transitions from increasing to decreasing or vice versa. By testing intervals around these points, we determined that the function increased on \((-\infty, -2)\) and \((0, \infty)\), and decreased on \((-2, 0)\). This analysis helps in identifying the local maxima and minima, valuable for understanding the general shape of the graph.

Concavity

Concavity gives us insights into how a function curves - whether it bends upwards or downwards. This is determined using the second derivative of the function. If the second derivative is positive, the function is concave up, like a cup; if negative, concave down, like a frown.

For the exercise's function \( h(x) \), we found the second derivative \( h''(x) = 20(x+1)^3 \). Examining where this second derivative equals zero gives us inflection points, indicating where concavity changes. Solving \( 20(x+1)^3 = 0 \) led us to have \( x = -1 \) as an inflection point.

Testing intervals around this point, we see between \( x < -1 \) the graph is concave down, and for \( x > -1 \), it is concave up. Such information is pivotal in sketching an accurate depiction of the function's graph.

Extrema

In calculus, extrema refer to the highest or lowest points on a graph within a specific interval. These are the local maximum and minimum points, which occur at critical points where a function changes from increasing to decreasing and vice versa.

From our interval testing using the first derivative results, we identified a local maximum at \( x = -2 \) where the function changes from increasing to decreasing, and a local minimum at \( x = 0 \) where it changes from decreasing to increasing.

Recognizing these extrema is crucial as they help highlight significant points in the graph where the function takes its highest and lowest values locally. They are key in understanding the overall behavior and plotting of the function, providing clear markers of turning points.