Problem 15 - University Calculus: Early Transcendentals

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y=(x-2)^{3}+1$$

Step-by-Step Solution

Answer

Inflection point at \( x = 2 \); no local extrema; function is increasing throughout.

Step 1: Find the First Derivative

To find critical points, determine where the first derivative equals zero or does not exist. Given the function \( y = (x-2)^3 + 1 \), find the derivative using the chain rule: \[ y' = 3(x-2)^2. \] This derivative exists everywhere, so set it to zero to find critical points: \[ 3(x-2)^2 = 0 \Rightarrow (x-2)^2 = 0 \Rightarrow x = 2. \] The critical point is at \( x = 2 \).

Step 2: Determine Local Extreme Points

Evaluate the second derivative to determine the concavity at the critical point. Compute the second derivative: \[ y'' = 6(x-2). \] Substitute \( x = 2 \): \( y''(2) = 0 \), which indicates neither concave up nor concave down. Hence, no local extreme points.

Step 3: Determine Inflection Points

An inflection point occurs where the second derivative changes sign. The second derivative is \( y'' = 6(x-2) \). Set \( y'' = 0 \) to find potential inflection points: \[ 6(x-2) = 0 \Rightarrow x = 2. \] Check sign changes around \( x = 2 \):- For \( x < 2 \), \( y'' < 0 \), indicating concave down.- For \( x > 2 \), \( y'' > 0 \), indicating concave up.Since there is a sign change, \( x = 2 \) is an inflection point.

Step 4: Graph the Function

Plot \( y = (x-2)^3 + 1 \).- Critical point at \( x = 2 \) with \( y = 1 \).- No local maxima or minima.- Inflection point at \( x = 2 \).- Analyze behavior: as \( x \to \infty \), \( y \to \infty \); as \( x \to -\infty \), \( y \to -\infty \).Draw a curve with an inflection point at \( x = 2 \), changing from concave down to concave up.

Key Concepts

Critical Points

In calculus, critical points are the values of the variable where the derivative of a function is zero or undefined. These points are significant because they can indicate where the function reaches a local maximum or minimum. For the given function \( y = (x-2)^3 + 1 \), we find its first derivative to locate critical points. By using the chain rule, the derivative is \( y' = 3(x-2)^2 \).
Setting this derivative to zero gives \( (x-2)^2 = 0 \), thus the critical point is at \( x = 2 \).

• This indicates a point where the slope of the function is zero.
• Check for local maxima, minima, or saddle points using further analysis.

By evaluating the second derivative, we determine the nature of this critical point.

Inflection Points

An inflection point on a curve occurs where the concavity of the function changes. It is identified by examining the second derivative of the function. For changes in concavity, the second derivative must change sign around the point of interest. For \( y = (x-2)^3 + 1 \), the second derivative is \( y'' = 6(x-2) \).
Setting \( y'' = 0 \) gives \( x = 2 \), suggesting a potential inflection point.

• For \( x < 2 \), \( y'' < 0 \) indicating concave down.
• For \( x > 2 \), \( y'' > 0 \) indicating concave up.
• The sign change confirms that \( x = 2 \) is indeed an inflection point.

This point marks a transition in the graph's shape, showcasing a switch from concave down to concave up.

Second Derivative

The second derivative of a function provides details about the curvature or concavity of the function's graph. It tells us how the slope of the function changes. For the function \( y = (x-2)^3 + 1 \), the second derivative is \( y'' = 6(x-2) \).
Calculating the second derivative helps us assess whether the point is a local extremum or an inflection point.

• If \( y'' > 0 \), the function is concave up (like a cup).
• If \( y'' < 0 \), the function is concave down (like a cap).
• If \( y'' = 0 \), further tests are required to determine the point's nature.

Here, \( y''(x) = 0 \) at \( x = 2 \) suggests a change in concavity, characteristic of an inflection point.

Concavity

Concavity describes the "bending" of a curve on a graph. It can show whether a function is curving upwards or downwards. Concave up looks like a bowl or a U, while concave down looks like an upside-down bowl.
Using the function \( y = (x-2)^3 + 1 \), we use the second derivative \( y'' = 6(x-2) \) to evaluate concavity.

• When \( y'' > 0 \), the function is concave up.
• When \( y'' < 0 \), the function is concave down.

At \( x = 2 \), the change from concave down to concave up indicates an inflection point. This detail is crucial when sketching graphs and can also imply potential optimization scenarios in practical applications.

Chain Rule

The chain rule is an essential calculus tool used for differentiating composite functions. A composite function is when one function is inside another, such as \( g(f(x)) \).
In our exercise, the function \( y = (x-2)^3 + 1 \) requires us to use the chain rule to find its derivative.

• The outer function is \( u^3 \).
• The inner function is \( (x-2) \).
• To differentiate, first find the derivative of the outer function: \( 3u^2 \).
• Then multiply by the derivative of the inner function: \( = 1 \).

Piecing it together gives \( y' = 3(x-2)^2 \). The chain rule simplifies the process of finding derivatives of nested functions effectively.