Problem 21 - 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. $$f(x)=3 x^{5}-5 x^{3}$$

Step-by-Step Solution

Answer

Critical points: \( x = 0, 1, -1 \).\nInflection points: \( x = 0, \pm \frac{1}{\sqrt{2}} \).\nLocal min at \( x = 1 \), max at \( x = -1 \).

Step 1: Find the First Derivative

To find the critical points, we first need to compute the first derivative of the function \( f(x) = 3x^5 - 5x^3 \). The derivative \( f'(x) \) is found using the power rule:\[ f'(x) = \frac{d}{dx}(3x^5) - \frac{d}{dx}(5x^3) = 15x^4 - 15x^2. \]

Step 2: Solve for Critical Points

Critical points occur where the first derivative is zero or undefined. Set \( f'(x) = 0 \):\[ 15x^4 - 15x^2 = 0. \]Factor out the greatest common factor:\[ 15x^2(x^2 - 1) = 0. \]This gives solutions:\( 15x^2 = 0 \) which solves to \( x = 0 \), and \( x^2 - 1 = 0 \) which solves to \( x = \pm 1 \). Thus, the critical points are \( x = 0, 1, -1 \).

Step 3: Find the Second Derivative

To determine the nature of critical points, compute the second derivative \( f''(x) \):\[ f''(x) = \frac{d}{dx}(15x^4 - 15x^2) = 60x^3 - 30x. \]

Step 4: Evaluate the Second Derivative at Critical Points

Evaluate \( f''(x) \) at each critical point:- For \( x = 0 \), \( f''(0) = 60(0)^3 - 30(0) = 0 \) (The test is inconclusive, use additional method to analyze).- For \( x = 1 \), \( f''(1) = 60(1)^3 - 30(1) = 30 \) (>0, local minimum).- For \( x = -1 \), \( f''(-1) = 60(-1)^3 - 30(-1) = -90 \) (<0, local maximum).

Step 5: Determine Inflection Points

Inflection points occur where \( f''(x) = 0 \). Set \( f''(x) = 0 \):\[ 60x^3 - 30x = 0. \]Factor:\[ 30x(2x^2 - 1) = 0. \]This gives solutions:\( 30x = 0 \) solves to \( x = 0 \), and \( 2x^2 - 1 = 0 \) solves to \( x = \pm \frac{1}{\sqrt{2}} \). These are the inflection points.

Step 6: Analyze Using a Graph

Use a graph of \( f(x) \) to further analyze each point:- At \( x = 0 \), the graph changes concavity, confirming an inflection point.- At \( x = 1 \), a local minimum occurs.- At \( x = -1 \), a local maximum occurs.These observations align with the second derivative test.

Key Concepts

Critical Points

In calculus, critical points are particularly special values where the derivative of a function, which indicates the rate of change, is either zero or undefined. For the function \( f(x) = 3x^5 - 5x^3 \), we first find the derivative:

• By applying the power rule, we differentiate to get \( f'(x) = 15x^4 - 15x^2 \).
• Setting \( f'(x) = 0 \) helps find critical points: \( 15x^4 - 15x^2 = 0 \).
• Factor out \( 15x^2 \), giving us \( 15x^2(x^2 - 1) = 0 \).
• This results in solutions \( x = 0, 1, -1 \) as critical points.

These points are essential to determine where the function might have local peaks or valleys.
Understanding critical points helps shape the function's graph, revealing both potential maxima and minima. They signal where significant changes in the behavior of the function might occur.

First Derivative

The first derivative of a function provides insights into the slope or the rate at which the function is changing at any given point. For our function \( f(x) = 3x^5 - 5x^3 \), the first derivative was found to be \( f'(x) = 15x^4 - 15x^2 \). Here’s how you understand its significance:

• The function increases where \( f'(x) > 0 \) and decreases where \( f'(x) < 0 \).
• At \( f'(x) = 0 \), the function is momentarily not changing—often leading to critical points.

This concept is the heart of calculus because it helps predict the increasing or decreasing nature of the function, making it foundational for finding critical points and analyzing the graph's contour.

Second Derivative

The second derivative is crucial in understanding the acceleration or concavity of the function. In our example, the second derivative is calculated as \( f''(x) = 60x^3 - 30x \).

• If \( f''(x) > 0 \), the function is concave up, resembling a "U" shape, indicating a local minimum.
• Conversely, if \( f''(x) < 0 \), the function is concave down, resembling an "n", indicating a local maximum.
• \( f''(x) = 0 \) suggests possible inflection points, where the concavity changes.

This analysis with the second derivative helps in determining how the slope from the first derivative is changing, which distinguishes between maximum and minimum critical points.

Local Maximum and Minimum

Identifying local maxima and minima is essential in understanding the overall landscape of a function. These points occur at critical points, which are tested using the second derivative:

• For \( x = 1 \), \( f''(1) = 30 \, (>0) \), indicating a local minimum.
• For \( x = -1 \), \( f''(-1) = -90 \, (<0) \), indicating a local maximum.
• Using the graph of \( f(x) \), these points are validated as the locations where the curve changes direction.

Local maxima and minima help us understand where the highest and lowest points in small local stretches of the function occur, which is valuable in optimization problems in real-world scenarios.

Inflection Points

Inflection points are where the function changes its concavity from up to down or vice versa. For the function at hand, these points are found by setting the second derivative to zero: \( f''(x) = 60x^3 - 30x = 0 \).

• This can be factored to \( 30x(2x^2 - 1) = 0 \), leading to \( x = 0, \pm \frac{1}{\sqrt{2}} \).
• These solutions are potential inflection points, confirmed by analyzing the changes in concavity on a graph.
• At an inflection point, the graph typically changes its curvature direction, characterizing the transition.

Inflection points are vital for understanding the transition of the curve, which affects how it will look in practice and helps in designing smooth transitions in engineering and graphics.