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Chapter 4: Problem 16

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y=1-(x+1)^{3}$$

Short Answer

Expert verified
The local and absolute extremum and inflection point is at (-1, 1).

Step by step solution

01

Identify the Function

The given function is \( y = 1 - (x+1)^3 \). This is a cubic function.
02

Find the First Derivative

To find the critical points, we first find the derivative of the function. The derivative is determined using the chain rule:\[y' = \frac{d}{dx}[1 - (x+1)^3] = -3(x+1)^2\]
03

Set the First Derivative to Zero

Find the critical points by solving \( y' = 0 \):\[-3(x+1)^2 = 0\]This simplifies to \[(x+1)^2 = 0\]Solving for \(x\), we get one solution:\[x = -1\]
04

Determine the Second Derivative

To find inflection points, calculate the second derivative:\[y'' = \frac{d}{dx}[-3(x+1)^2] = -6(x+1)\]
05

Set the Second Derivative to Zero

Solve for potential inflection points by setting \( y'' = 0 \):\[-6(x+1) = 0\]Solving for \(x\), we have:\[x = -1\]
06

Evaluate the Coordinates of Critical and Inflection Points

Substitute \(x = -1\) into the original equation to find the y-coordinate:\[y = 1 - (-1+1)^3 = 1 - 0 = 1\]So the coordinates of both the local/absolute extremum and inflection point are \((-1, 1)\).
07

Graphing the Function

The graph of the function \( y = 1 - (x+1)^3 \) is a downward-shifted cubic curve. The point \((-1, 1)\) is both an extremum and an inflection point, meaning the graph changes concavity at this point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Critical Points
Critical points are essential in understanding a function's behavior, especially for identifying the locations of maximum or minimum values. To find critical points of a function, we perform the following steps:

  • First, calculate the first derivative of the function. The first derivative, denoted as \(y'\), helps us determine the rate of change of the function.
  • Set the first derivative equal to zero. Solving \(y' = 0\) enables us to find the values of \(x\) where the slope of the tangent to the function is zero. These are the candidate points for maximum or minimum values.
  • In the case of \(y = 1 - (x+1)^3\), the first derivative is \(y' = -3(x+1)^2\). Setting this to zero, we only find one critical point at \(x = -1\). This means \(x = -1\) is potentially where the function may have a "peak" or "valley".
It's important to further investigate each critical point to determine whether it's a local maximum, local minimum, or a saddle point.
Inflection Points
Inflection points are where a function changes its concavity, which means it transitions from concave up to concave down or vice versa. The steps to find inflection points are as follows:

  • Calculate the second derivative of the function. The second derivative, denoted as \(y''\), helps us understand how the rate of change itself changes.
  • Set the second derivative equal to zero to find possible inflection points. Solving \(y'' = 0\) gives the values of \(x\) where inflection points can occur.
  • For the function \(y = 1 - (x+1)^3\), we find \(y'' = -6(x+1)\), which equals zero at \(x = -1\). This implies the function might change concavity at this point.
However, it's necessary to determine whether there's indeed a change in concavity at \(x = -1\) by analyzing the behavior on either side of this point.
Derivatives
Understanding derivatives is crucial for analyzing cubic functions, as they help us explore the function's behavior.

The **first derivative** \(y'\) of a function allows us to examine the function's slope, indicating where it rises or falls:

  • Using the chain rule, the derivative of \(y = 1 - (x+1)^3\) becomes \(y' = -3(x+1)^2\).
  • This provides us with a tool to find points where the function's rate of change is zero (i.e., critical points).
The **second derivative** \(y''\) provides insight into the concavity of the function:

  • The second derivative is calculated from the first derivative. Here, \(y'' = -6(x+1)\), which helps us identify possible inflection points.
  • It tells us where the graph changes its concavity and further reveals the function's complexity.
Understanding both derivatives assists in thoroughly analyzing and graphing cubic functions.
Graphing Functions
Graphing functions is a visual approach to understanding how mathematical equations behave in different intervals.

To graph a function:

  • Identify critical points and inflection points. Here, for \(y = 1 - (x+1)^3\), critical and inflection points are both at \((-1, 1)\).
  • Evaluate the end behavior. For our cubic function, as \(x\) approaches infinity, the \(y\)-values decrease due to the \(-x^3\) nature of the expression.
  • Sketch the curve, especially noting the behavior around critical points and inflection points. Since \((-1, 1)\) acts both as an extremum and an inflection point, the graph transitions between concave up and concave down here.
The visual representation of the function allows students to correlate algebraic computations with their graphical interpretations, deepening their understanding.

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Most popular questions from this chapter

Find the critical points, domain endpoints, and extreme values (absolute and local) for each function. $$y=\left\\{\begin{array}{ll}-\frac{1}{4} x^{2}-\frac{1}{2} x+\frac{15}{4}, & x \leq 1 \\\x^{3}-6 x^{2}+8 x, & x>1\end{array}\right.$$

Solve the initial value problems in Exercises. $$\frac{d v}{d t}=\frac{3}{t \sqrt{t^{2}-1}}, \quad t>1, v(2)=0$$

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