Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the critical points and use the test of your choice to decide which critical points give a local maximum value and which give a local minimum value. What are these local maximum and minimum values? $$ f(x)=(x-2)^{5} $$

Step-by-Step Solution

Answer

The function has a critical point at \( x=2 \), but it has neither a local maximum nor minimum there.

Step 1: Find the first derivative

To find the critical points, we first need to find the first derivative of the function. Given the function \( f(x) = (x-2)^5 \), we use the power rule to find that: \[ f'(x) = 5(x-2)^4 \].

Step 2: Set the first derivative equal to zero

Critical points occur where the first derivative is zero or undefined. Since \( f'(x) = 5(x-2)^4 \) is defined for all real \( x \), we set the derivative equal to zero to find critical points: \[ 5(x-2)^4 = 0 \].

Step 3: Solve for \( x \)

Solving \( 5(x-2)^4 = 0 \), we divide both sides by 5, obtaining \((x-2)^4 = 0\). Solving for \( x \), we find \( x=2 \) is the only solution.

Step 4: Second derivative test for concavity

To determine whether \( x=2 \) is a local maximum or minimum, use the second derivative test. First, find the second derivative: \[ f''(x) = 20(x-2)^3 \]. Evaluate at \( x = 2 \): \( f''(2) = 20(2-2)^3 = 0 \).

Step 5: Test using first derivative

Since the second derivative test is inconclusive (\( f''(2) = 0 \)), check the first derivative around \( x=2 \). For values close to 2, such as for an \( x < 2 \) and \( x > 2 \), observe that \( f'(x) > 0 \,.(\ (x-2)^4 \) is always non-negative). Thus, \( f'(x) \) remains positive, meaning \( x=2 \) is neither a local maximum nor minimum but instead a point of inflection.

Key Concepts

First Derivative

The first derivative helps us find critical points of a function. These occur where the slope of the function is zero or undefined. For our function, \( f(x) = (x-2)^5 \), the first derivative is found using the Power Rule. The Power Rule states that if you have a function of the form \( x^n \), its derivative will be \( nx^{n-1} \). For \( (x-2)^5 \), applying the Power Rule gives \( f'(x) = 5(x-2)^4 \).

Once we have the first derivative, we set it to zero to find the critical points. In our case, this results in \( 5(x-2)^4 = 0 \). Solving \( (x-2)^4 = 0 \) leads us to the critical point \( x=2 \).

Critical points are vital in determining the behavior of a function at certain values. They can indicate potential local maxima or minima, or even points of inflection.

Second Derivative Test

The second derivative test is a method used to classify critical points as local minima, maxima, or points of inflection. It evaluates the concavity of a function at its critical points. For our function \( f(x) = (x-2)^5 \), the second derivative is \( f''(x) = 20(x-2)^3 \).

We computed \( f''(2) = 20(2-2)^3 = 0 \). When the second derivative equals zero, the test is inconclusive. This means \( x=2 \) does not fit neatly into the category of a local minimum or maximum based on the second derivative alone.

In such cases where the outcome is zero, returning to the first derivative analysis can help further understand the behavior of the function near the critical point.

Point of Inflection

A point of inflection is where the concavity of a function changes. It can occur when the second derivative is zero or undefined, but it's important to further investigate. In our example, \( f''(2) \) equaled zero, a classic sign of a potential point of inflection.

By analyzing the behavior of the first derivative around \( x=2 \), we found \( f'(x) > 0 \) for \( x < 2 \) and \( x > 2 \). Typically, if the function moves from concave up to concave down, or vice versa, a point of inflection is present. However, if the function doesn't change signs and stays positive or negative, it confirms the point at \( x=2 \) doesn't correspond to a local maximum or minimum, solidifying it as a point of inflection.

Power Rule

The Power Rule is a core concept of calculus, providing a straightforward way to find derivatives of power functions. As mentioned earlier, for any function of the form \( x^n \), the derivative is \( nx^{n-1} \). This simplifies the differentiation process significantly.

In practice, \( f(x) = (x-2)^5 \) required the Power Rule to determine its first derivative, \( f'(x) = 5(x-2)^4 \). This shows the Power Rule's invaluable role in calculus for quickly approaching critical points and understanding function behaviors.

Learning to apply the Power Rule effectively allows for ease in exploring deeper concepts like optimization problems—skills that are crucial for analyzing and understanding functions comprehensively.