Problem 10 - Calculus with Concepts in Calculus

Question

Use the First Derivative Test to determine the relative extreme values (if any) of the function. $$ f(x)=x^{3}-12 x+2 $$

Step-by-Step Solution

Answer

Relative maximum at \( x = -2 \) and relative minimum at \( x = 2 \).

Step 1: Find the First Derivative

The first step in the First Derivative Test is to compute the first derivative of the function. For the given function \( f(x) = x^3 - 12x + 2 \), use the power rule to find \( f'(x) \).The derivative is: \[ f'(x) = 3x^2 - 12 \]

Step 2: Solve for Critical Points

To find critical points, set the first derivative \( f'(x) = 3x^2 - 12 \) to zero and solve for \( x \).\[ 3x^2 - 12 = 0 \]Factor the equation:\[ 3(x^2 - 4) = 0 \]So, \( x^2 - 4 = 0 \).\[ x^2 = 4 \]\[ x = \pm 2 \]Thus, the critical points are \( x = 2 \) and \( x = -2 \).

Step 3: Perform the First Derivative Test

Use the First Derivative Test to classify each critical point. This involves evaluating the sign of \( f'(x) \) on intervals defined by the critical points.- Interval \((-\infty, -2)\): Choose \( x = -3 \):\[ f'(-3) = 3(-3)^2 - 12 = 27 - 12 = 15 \] (positive)- Interval \((-2, 2)\): Choose \( x = 0 \):\[ f'(0) = 3(0)^2 - 12 = -12 \] (negative)- Interval \((2, \infty)\): Choose \( x = 3 \):\[ f'(3) = 3(3)^2 - 12 = 27 - 12 = 15 \] (positive)The sign changes from positive to negative at \( x = -2 \) and from negative to positive at \( x = 2 \).

Step 4: Determine Relative Extrema

Based on the First Derivative Test results:- At \( x = -2 \), \( f'(x) \) changes from positive (increasing) to negative (decreasing), indicating a relative maximum.- At \( x = 2 \), \( f'(x) \) changes from negative (decreasing) to positive (increasing), indicating a relative minimum.

Key Concepts

Understanding Critical Points

In calculus, critical points are crucial for identifying where a function might have potential highs or lows. To find these critical points, you need to look at where the derivative of a function equals zero or where it becomes undefined. Specifically, if you have a function \( f(x) \), the critical points are the \( x \)-values for which \( f'(x) = 0 \) or \( f'(x) \) does not exist.

• These are the points at which the tangent to the curve is horizontal (slope zero), resembling a flat spot on the graph.
• Remember, a critical point doesn't necessarily mean a maximum or minimum exists there; it simply indicates a potential for one.
• In the function \( f(x) = x^3 - 12x + 2 \), we determined the critical points by setting \( f'(x) = 3x^2 - 12 \) to zero, resulting in \( x = 2 \) and \( x = -2 \).

Breaking down the derivative and solving it is often the trickiest part, but it's foundational in exploring the nature of these important points.

Exploring Relative Extrema

Relative extrema refer to the local high and low points of a function within a certain interval. These are effectively the peaks and valleys on a graph when viewed over a specific range.

• A **relative maximum** is a point on the graph where the function changes from increasing to decreasing. It's essentially a peak, or a hilltop, in the local interval.
• A **relative minimum** is where the function changes from decreasing to increasing. Think of it as a local valley.
• Using the First Derivative Test, we can determine the nature of relative extrema. For \( f(x) = x^3 - 12x + 2 \), it was evident when \( x = -2 \) that the function had a relative maximum, while at \( x = 2 \), it had a relative minimum.

These points provide valuable insights for analyzing the function's behavior in calculus problem solving.

Mastering Calculus Problem Solving with the First Derivative Test

The First Derivative Test is a powerful tool in problem-solving and understanding the behavior of functions in calculus. It involves analyzing the derivative to determine where the function is increasing or decreasing. Here’s a breakdown of the process:

• First, compute the first derivative of the function to find the slope formula.
• Next, find critical points by setting the derivative equal to zero and solving for the variable.
• With the critical points, divide the number line into intervals and evaluate the sign of the derivative (positive or negative) in those intervals.
• Observe the sign changes across those intervals to classify the nature of the critical points—whether they are points of relative maximum or minimum.

Tips for Success: Always check the intervals carefully, ensuring to pick test points that give clear insights. Also, remember that critical points could fail to be extrema if the sign doesn't change. In our exercise with \( f(x) = x^3 - 12x + 2 \), the test confirmed the nature of the critical points as either a maximum or minimum, showcasing the test's effectiveness.