Problem 25 - Calculus of a Single Variable

Question

Applying the First Derivative Test In Exercises \(17-40\) , (a) find the critical numbers of \(f\) (if any), (b) find the open interval(s) on which the function is increasing or decreasing, (c) apply the First Derivative Test to identify all relative extrema, and (d) use a graphing utility to confirm your results. $$ f(x)=\frac{x^{5}-5 x}{5} $$

Step-by-Step Solution

Answer

The function has critical numbers at \(x = ±1\). It is increasing on \((-∞, -1)\) and \((1, ∞)\), and decreasing on \((-1, 1)\). The function has a local maximum at \(x = -1\) and a local minimum at \(x = 1\). The graph of the function confirms these results.

Step 1: Calculating Derivative

Using the power rule, the derivative of \(f(x)\) is \(f'(x) = x^4 - 1\)

Step 2: Finding Critical Numbers

Critical numbers are where the derivative is zero or undefined. For this exercise, the derivative \(f'(x) = x^4 - 1\) exists for all \(x\), but it's zero when \(x^4 - 1 = 0\), or \(x = ±1\). So, ±1 are the critical numbers.

Step 3: Determining Increasing and Decreasing Intervals

We can use the sign test for the intervals \((-∞, -1)\), \((-1, 1)\), and \((1, ∞)\). Pick a test number from each interval and check the sign of \(f'(x)\):1. Interval \((-∞, -1)\): Choose \(x = -2\), then \(f'(-2) = 16 - 1 = 15 > 0\). So, \(f(x)\) is increasing in the interval \((-∞, -1)\).2. Interval \((-1, 1)\): Choose \(x = 0\), then \(f'(0) = -1 < 0\). So, \(f(x)\) is decreasing in the interval \((-1, 1)\).3. Interval \((1, ∞)\): Choose \(x = 2\), then \(f'(2) = 16 - 1 = 15 > 0\). So, \(f(x)\) is increasing in the interval \((1, ∞)\).

Step 4: Applying the First Derivative Test

Since \(f(x)\) changed from increasing to decreasing at \(x = -1\), \(f(x)\) has a local maximum at \(x = -1\). Likewise, it changed from decreasing to increasing at \(x = 1\), so \(f(x)\) has a local minimum at \(x = 1\). These are classified as relative extrema.

Step 5: Confirming Results with Graph

When the function \(f(x) = \frac{{x^5 - 5x}}{5}\) is graphed, it shows a peak at \(x = -1\) and a valley at \(x = 1\), which confirms our results.

Key Concepts

Critical Numbers

Critical numbers are values of the independent variable, often denoted as 'x', where the derivative of a function equals zero or does not exist. These points are essential in determining the local behavior of the function, such as where it may reach maximum or minimum values, or where the slope of the tangent to the function's graph is horizontal. In our exercise, calculating the derivative of the function \( f(x) = \frac{x^{5}-5x}{5} \) and setting it equal to zero, gives us \( f'(x) = x^4 - 1 \). Solving for \( x \) where \( f'(x) = 0 \) leads to \( x = \pm1 \), indicating that \( x = -1 \) and \( x = 1 \) are the critical numbers for this particular function.

Identifying critical numbers is a fundamental step in analyzing the overall trend and turning points of the function, which are crucial for understanding its graph and various applications.

Increasing and Decreasing Intervals

A function is said to be increasing on an interval if, as x moves from left to right, the function's values become larger. Conversely, a function is decreasing on an interval if the values become smaller. To determine the increasing and decreasing intervals for a function, we analyze the sign of its first derivative. If \( f'(x) > 0 \) over an interval, then \( f(x) \) is increasing on that interval. If \( f'(x) < 0 \), then \( f(x) \) is decreasing.

In our exercise, we used test points to determine where the function \( f(x) = \frac{x^{5}-5x}{5} \) was increasing or decreasing. For the intervals \( (-\bullet, -1) \), \( (-1, 1) \), and \( (1, \bullet) \), we found that \( f(x) \) is increasing when \( x < -1 \) and \( x > 1 \), but decreasing when \( -1 < x < 1 \). It's also important to note that an increasing or decreasing trend leads us directly into the concept of relative extrema - the peaks and valleys of the function's graph.

Relative Extrema

Relative extrema refer to the local maximums and minimums of a function. A relative maximum is a point where a function's value is higher than all other values in some interval surrounding that point. Similarly, a relative minimum is a point where the function's value is lower than all other values in the surrounding interval. These points are often located at critical numbers where the function's derivative changes sign.

In the context of the provided exercise, we have seen that after applying the First Derivative Test, a local maximum is identified at \( x = -1 \), due to the fact that \( f(x) \) changed from increasing to decreasing at this critical number. Conversely, a local minimum is located at \( x = 1 \), because there the function changes from decreasing to increasing. Graphing the function confirms these findings visually, as we see a peak (maximum) at \( x = -1 \) and a valley (minimum) at \( x = 1 \). Understanding these concepts can help students to analyze functions and their graphs more thoroughly, identifying not just where the function climbs or descends, but also where it reaches its most significant or least significant points within certain ranges.