Problem 13 - Applied Calculus

Question

Use the first derivative to find all critical points and use the second derivative to find all inflection points. Use a graph to identify each critical point as a local maximum, a local minimum, or neither. $$f(x)=x^{3}-3 x+10$$

Step-by-Step Solution

Answer

Critical points: \(x = -1\) (max), \(x = 1\) (min). Inflection point: \(x = 0\).

Step 1: Find the first derivative

To begin finding critical points, calculate the first derivative of the function \(f(x)=x^3-3x+10\). The first derivative is given by: \[ f'(x) = \frac{d}{dx}(x^3 - 3x + 10) = 3x^2 - 3. \]

Step 2: Set the first derivative to zero

Critical points occur where the derivative is zero or undefined. Set the first derivative \(3x^2 - 3\) equal to zero and solve for \(x\): \[ 3x^2 - 3 = 0. \] Simplifying, we have: \[ x^2 = 1. \] Thus, \(x = \pm 1\).

Step 3:

The critical points are \(x = 1\) and \(x = -1\).

Step 3: Use the second derivative to classify critical points

Calculate the second derivative of the function \(f(x)\): \[ f''(x) = \frac{d}{dx}(3x^2 - 3) = 6x. \] Substitute \(x = 1\) and \(x = -1\) into the second derivative to determine concavity: - For \(x = 1\), \(f''(1) = 6(1) = 6\), which is positive, indicating a local minimum. - For \(x = -1\), \(f''(-1) = 6(-1) = -6\), which is negative, indicating a local maximum.

Step 4: Find the second derivative to determine inflection points

An inflection point occurs where the second derivative changes sign. Set \(f''(x) = 6x = 0\). Solving for \(x\) gives \(x = 0\). Substitute values around \(x = 0\) to confirm the change in sign: - For \(x < 0\), \(f''(x) < 0\). - For \(x > 0\), \(f''(x) > 0\). This sign change confirms an inflection point at \(x = 0\).

Step 5: Graph the function to confirm critical points

Graph the function \(f(x) = x^3 - 3x + 10\). The graph should confirm: - A local minimum at \(x = 1\) (\(f(x)\) changes from decreasing to increasing). - A local maximum at \(x = -1\) (\(f(x)\) changes from increasing to decreasing). - An inflection point at \(x = 0\) (concavity changes from downward to upward).

Key Concepts

First Derivative

The first derivative of a function gives us valuable information about the slope of the tangent line at any given point on the function's graph. For the function \( f(x) = x^3 - 3x + 10 \), the first derivative is calculated to be \( f'(x) = 3x^2 - 3 \). This derivative tells us how the function is increasing or decreasing at each point.

• The critical points are found by setting the first derivative \( f'(x) \) equal to zero. This gives us the equations where the slope of the tangent is horizontal, indicating possible peaks, troughs, or points of inflection.
• For this particular function, solving \( 3x^2 - 3 = 0 \) yields \( x = 1 \) and \( x = -1 \) as critical points.

Determining these points is the first step in analyzing the behavior of the function.

Second Derivative

The second derivative provides insight into the concavity of a function, indicating whether it curves upwards or downwards at a given point. For \( f(x) = x^3 - 3x + 10 \), the second derivative \( f''(x) = 6x \) helps us classify critical points further.

• A positive second derivative, such as \( f''(1) = 6 \), suggests the function is concave up at \( x = 1 \), indicating a local minimum.
• A negative second derivative, like \( f''(-1) = -6 \), indicates the function is concave down at \( x = -1 \), suggesting a local maximum.

This information helps us determine how the function behaves around critical points.

Inflection Points

Inflection points occur where a function changes concavity, from concave up to concave down or vice versa. These points are found by examining where the second derivative changes sign. For \( f(x) = x^3 - 3x + 10 \), setting the second derivative \( f''(x) = 6x \) equal to zero identifies \( x = 0 \) as a potential inflection point.
To confirm:

• Check values just below and above \( x = 0 \) to detect a sign change in \( f''(x) \).
• If \( f''(x) \) changes from negative to positive, the concavity transitions from downwards to upwards, confirming an inflection point at \( x = 0 \).

Identifying inflection points adds a deeper understanding of the function's graph.

Local Maximum

A local maximum is where a function reaches a peak value in a certain region, higher than any other nearby values. At \( x = -1 \), the function \( f(x) = x^3 - 3x + 10 \) has a local maximum.

• This is determined both by setting \( f'(x) = 0 \) and confirming with the second derivative test.
• Since \( f''(-1) = -6 \) is less than zero, it indicates concave down, making \( x = -1 \) a local maximum.
The function decreases before and increases after this point, confirming it's a peak in the local region.

Local Minimum

A local minimum occurs at a point where the function has a dip, lower than any nearby values. For the function \( f(x) = x^3 - 3x + 10 \), \( x = 1 \) is a local minimum.

• First, we set \( f'(x) = 0 \) to identify possible minima.
• The second derivative, \( f''(1) = 6 \), is positive, indicating concave up at this point.

This suggests \( x = 1 \) is where the function reaches a trough, with values rising before and after, marking it clearly as a local minimum.