Problem 42 - Calculus with Concepts in Calculus

Question

Use the First Derivative Test or the Second Derivative Test to determine the relative extreme values, if any, of the function. Then sketch the graph of the function. $$ f(x)=x^{5}-5 x $$

Step-by-Step Solution

Answer

Relative maximum at \(x = -1\) and relative minimum at \(x = 1\).

Step 1: Find the First Derivative

To apply the First Derivative Test, first find the derivative of the function \( f(x) = x^5 - 5x \). The derivative is found using the power rule. \[ f'(x) = 5x^4 - 5 \]

Step 2: Identify Critical Points

Critical points occur where the derivative is zero or undefined. Solve \( f'(x) = 0 \) to find these points. \[ 5x^4 - 5 = 0 \] \[ x^4 = 1 \] \[ x = \pm 1 \] Thus, the critical points are \( x = 1 \) and \( x = -1 \).

Step 3: First Derivative Test for Increasing/Decreasing Intervals

Use the First Derivative Test by testing intervals around the critical points. Choose test points from the intervals: \((-\infty, -1)\), \((-1, 1)\), and \((1, \infty)\). Evaluate the sign of \( f'(x) \). - For \((-\infty, -1)\), choose \(x = -2\): \( f'(-2) = 5(-2)^4 - 5 = 75 \) (positive)- For \((-1, 1)\), choose \(x = 0\): \( f'(0) = 5(0)^4 - 5 = -5 \) (negative) - For \((1, \infty)\), choose \(x = 2\): \( f'(2) = 5(2)^4 - 5 = 75 \) (positive) Thus, the function is increasing on \((-\infty, -1)\) and \((1, \infty)\), and decreasing on \((-1, 1)\).

Step 4: Determine Relative Extrema Using the First Derivative Test

Using the information about increasing and decreasing intervals:- At \(x = -1\), the function changes from increasing to decreasing, indicating a relative maximum.- At \(x = 1\), the function changes from decreasing to increasing, indicating a relative minimum.Thus, \(f(x)\) has a relative maximum at \(x = -1\) and a relative minimum at \(x = 1\).

Step 5: Sketch the Graph

Based on the critical points and relative extrema:- The function has a maximum at \(x = -1\) and a minimum at \(x = 1\).- It increases as \(x\) approaches \(-1\) from the left and decreases between \(-1\) and \(1\).- The function increases again for \(x > 1\). Plot these key features on a graph to sketch the overall shape of the function.

Key Concepts

Critical Points

Critical points are the values of \( x \) where a function's derivative is zero or undefined. Finding critical points is crucial because they help us identify where a function may change its behavior from increasing to decreasing, or vice versa.
To find the critical points of a given function, say \( f(x) = x^5 - 5x \), we need to first compute its derivative. For this function, applying the power rule, we get \( f'(x) = 5x^4 - 5 \).
Next, we set the derivative equal to zero to solve for \( x \):

• \( 5x^4 - 5 = 0 \)
• \( x^4 = 1 \)

This leads to the critical points \( x = 1 \) and \( x = -1 \). These points are essential for understanding the nature of the function around these values.

Relative Extrema

Relative extrema refer to the local maximum or minimum points in a function. These are points where the function transitions its direction, identified using the first derivative. At these critical points, the function can either have a peak (maximum) or a trough (minimum).
Using the First Derivative Test, we can determine the relative extrema of \( f(x) = x^5 - 5x \) by observing which direction the function moves at and around the critical points.

• For \( x = -1 \), the function shifts from an increasing slope to a decreasing slope, causing \( x = -1 \) to be a relative maximum.
• For \( x = 1 \), the function goes from decreasing to increasing, resulting in a relative minimum at \( x = 1 \).

These localized high and low points give us a picture of the "hills and valleys" in the function's graph.

Increasing and Decreasing Intervals

Identifying where a function is increasing or decreasing gives insightful information about its behavior across different regions. This can be determined by looking at the sign of the derivative in various intervals around the critical points.
For the function \( f(x) = x^5 - 5x \), we checked intervals with test points based on our critical points \( x = -1 \) and \( x = 1 \).

• Between \((-\infty, -1)\): Check \( x = -2 \), and since \( f'(-2) = 75 \), the sign is positive, meaning the function is increasing.
• Between \((-1, 1)\): Check \( x = 0 \), where \( f'(0) = -5 \), indicating a negative sign and that the function is decreasing.
• Between \((1, \infty)\): Check \( x = 2 \), giving \( f'(2) = 75 \) again positive, showing the function is increasing once more.

So, the function increases on \((-\infty, -1)\) and \((1, \infty)\), but decreases on \((-1, 1)\). This analysis helps sketch the overall trend and shape of the function.

Second Derivative Test

The Second Derivative Test is another method to determine the nature of critical points and ascertain whether they are maxima, minima, or points of inflection. It involves looking at the second derivative of the function.
For \( f(x) = x^5 - 5x \), the second derivative is found by differentiating \( f'(x) = 5x^4 - 5 \) once more. This results in the second derivative \( f''(x) = 20x^3 \).
Here's how it helps:

• If \( f''(x) \) is positive at a critical point, the function has a relative minimum there.
• If \( f''(x) \) is negative, the function has a relative maximum at that point.
• If \( f''(x) = 0 \), the test is inconclusive, and further analysis might be necessary.

Overall, the Second Derivative Test offers a more extensive perspective on the concavity of the graph and helps confirm the findings from other methods.