Problem 54 - University Calculus: Early Transcendentals

Question

Find the extreme values (absolute and local) of the function over its natural domain, and where they occur. $$y=x^{3}-2 x+4$$

Step-by-Step Solution

Answer

Critical points occur at \( x = \sqrt[3]{2} \) and \( x = -\sqrt[3]{2} \); both are inflection points as \( y'' = 6x = 0 \) at \( x = 0 \).

Step 1: Understand the Problem

We need to find both the absolute and local extreme values (maxima and minima) of the function \( y = x^3 - 2x + 4 \), and determine where they occur.

Step 2: Find the First Derivative

To find critical points where extreme values occur, we first find the derivative of the function with respect to \( x \). Let's find \( y' \).

Step 3: Solve for Critical Points

Set the first derivative equal to zero and solve for \( x \) to find critical points. If \( y' = 0 \), it reveals potential points for examining extreme values.

Step 4: Find the Second Derivative

Determine if critical points are maxima or minima by finding the second derivative \( y'' \). Use it to apply the second derivative test.

Step 5: Test Critical Points

Evaluate \( y'' \) at each critical point. If \( y'' > 0 \), the point is a local minimum; if \( y'' < 0 \), the point is a local maximum. If \( y'' = 0 \), the test is inconclusive.

Step 6: Determine Extreme Values

Substitute each critical point \( x \) back into the original function \( y \) to find the respective extreme values.

Key Concepts

Derivatives

Derivatives are a fundamental concept in calculus, representing the rate at which a function changes. For a given function like \( y = x^3 - 2x + 4 \), finding the derivative helps us understand its behavior. The derivative, expressed as \( y' \) for this function, helps identify where the function is increasing or decreasing. To find the first derivative, apply the power rule: the derivative of \( x^n \) is \( nx^{n-1} \). Therefore, for \( y = x^3 - 2x + 4 \):

• The derivative of \( x^3 \) is \( 3x^2 \).
• The derivative of \(-2x\) is \(-2\).
• The derivative of the constant 4 is 0.

Thus, the first derivative or \( y' = 3x^2 - 2 \). This equation provides insight into critical points and is essential for finding extrema.

Critical Points

Critical points occur where the derivative of a function is zero or undefined. These points are potential candidates for extreme values—where the function could have local maxima or minima. To find critical points for the function \( y = x^3 - 2x + 4 \), set its derivative \( y' = 3x^2 - 2 \) to zero:

• \( 3x^2 - 2 = 0 \)
• Solve for \( x \): \( x = \pm \sqrt{\frac{2}{3}} \)

There are two critical points: \( x = \sqrt{\frac{2}{3}} \) and \( x = -\sqrt{\frac{2}{3}} \). These are the points where the function may reach its local extrema. Investigating the behavior of the function at these critical points is crucial for determining the type of extremum.

Second Derivative Test

The second derivative test provides a way to classify critical points found from the first derivative. It determines if each critical point is a local maximum, minimum, or inconclusive. First, find the second derivative \( y'' \) of the function \( y = x^3 - 2x + 4 \):

• Differentiating \( y' = 3x^2 - 2 \) yields \( y'' = 6x \).

Now, evaluate \( y'' \) at each critical point:

• At \( x = \sqrt{\frac{2}{3}} \), \( y'' = 6\sqrt{\frac{2}{3}} > 0 \); indicates a local minimum.
• At \( x = -\sqrt{\frac{2}{3}} \), \( y'' = -6\sqrt{\frac{2}{3}} < 0 \); indicates a local maximum.

This test helps confirm the nature of each extremum at the critical points.

Local and Absolute Extrema

Local extrema occur at critical points where the function reaches a relative peak (maximum) or trough (minimum) compared to nearby points. Absolute extrema represent the highest or lowest value over the entire domain of the function. To find them, evaluate the original function at critical points. For \( y = x^3 - 2x + 4 \):

• At \( x = \sqrt{\frac{2}{3}} \), \( y = \left(\sqrt{\frac{2}{3}}\right)^3 - 2\left(\sqrt{\frac{2}{3}}\right) + 4 \).
• At \( x = -\sqrt{\frac{2}{3}} \), \( y = \left(-\sqrt{\frac{2}{3}}\right)^3 - 2\left(-\sqrt{\frac{2}{3}}\right) + 4 \).

These calculations help find local extrema. The context of the function—whether it's considered over a bounded interval or its entire domain—influences if these extrema are absolute, meaning they are the highest or lowest in the entire range.