Problem 18 - 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. \(f(x)=x^{4}-4 x^{3}+10\)

Step-by-Step Solution

Answer

Critical points are at \(x = 0\) and \(x = 3\); local minimum at \(x = 3\). Inflection point at \(x = 2\).

Step 1: Find the First Derivative

Start by differentiating the function \(f(x) = x^4 - 4x^3 + 10\). The first derivative \(f'(x)\) is found using the power rule for each term:\[ f'(x) = 4x^3 - 12x^2 \].

Step 2: Find Critical Points

Critical points occur when the first derivative is zero or undefined. Set the first derivative equal to zero and solve for \(x\):\[ 4x^3 - 12x^2 = 0 \].Factor the equation:\[ 4x^2(x - 3) = 0 \].Thus, the critical points are \(x = 0\) and \(x = 3\).

Step 3: Verify Critical Points Using Second Derivative

To confirm whether the critical points are local maxima, minima, or neither, use the second derivative test. First, find the second derivative of \(f(x)\):\[ f''(x) = 12x^2 - 24x \].Evaluate \(f''(x)\) at each critical point:- For \(x = 0\), \(f''(0) = 12(0)^2 - 24(0) = 0\).- For \(x = 3\), \(f''(3) = 12(3)^2 - 24(3) = 36 > 0\).

Step 4: Analyze Second Derivative Results

Using the second derivative test:- At \(x = 0\), since \(f''(0) = 0\), the second derivative test is inconclusive; check using a graph or another method.- At \(x = 3\), since \(f''(3) > 0\), the function has a local minimum.

Step 5: Determine Inflection Points

Inflection points occur where the second derivative changes sign. Set \(f''(x)\) to zero and solve for \(x\):\[ 12x^2 - 24x = 0 \].Factor the equation:\[ 12x(x - 2) = 0 \].Thus, potential inflection points are \(x = 0\) and \(x = 2\).

Step 6: Confirm Inflection Points

Test intervals around \(x = 0\) and \(x = 2\) to see where \(f''(x)\) changes sign:- Test points around \(x = 0\), e.g., \(x = -1\) and \(x = 1\).- Test points around \(x = 2\), e.g., \(x = 1\) and \(x = 3\).For \(x = 0\), \(f''(x)\) does not change sign.For \(x = 2\), \(f''(x)\) changes from negative to positive, indicating an inflection point.

Step 7: Final Step: Graph Analysis

Graph \(f(x)\) to visually confirm the behavior around critical and inflection points.- At \(x = 0\), verify with graph this is a saddle point or neither.- At \(x = 3\), confirm it is a local minimum.- At \(x = 2\), see the change in concavity which indicates an inflection point.

Key Concepts

First Derivative

The first derivative of a function, denoted as \( f'(x) \), plays a crucial role in finding critical points. These points are where the function's slope is either zero or undefined. Finding the first derivative involves using differentiation rules, such as the power rule for each term in the function's equation.

For the function \( f(x) = x^4 - 4x^3 + 10 \), the first derivative is calculated as \( f'(x) = 4x^3 - 12x^2 \). The process involves differing each term by reducing the power by one and multiplying the original coefficient by the power. For example, the derivative of \( x^4 \) is \( 4x^3 \).

Once the first derivative is determined, setting it to zero helps find critical points since that's where the graph's slope is horizontal. In this case, solving \( 4x^3 - 12x^2 = 0 \) leads to \( x = 0 \) and \( x = 3 \). These are your critical points, providing possible hills or valleys of the graph.

Second Derivative

The second derivative, expressed as \( f''(x) \), is instrumental in determining the concavity of the function and analyzing the nature of critical points. By assessing the second derivative, you can apply the second derivative test to provide more information about critical points found from the first derivative.

The second derivative is the derivative of the first derivative. For \( f'(x) = 4x^3 - 12x^2 \), the second derivative is \( f''(x) = 12x^2 - 24x \). This step requires differentiating each term of the first derivative again.

When evaluating the second derivative at critical points, you can classify them based on the result:

• If \( f''(x) > 0 \), the function is concaving upwards, suggesting a local minimum.
• If \( f''(x) < 0 \), it indicates concaving downwards, hinting at a local maximum.
• If \( f''(x) = 0 \), the test is inconclusive, and other methods might be needed.

In this example, evaluating \( f''(3) = 36 \) reveals a local minimum at \( x = 3 \), while \( f''(0) = 0 \) is inconclusive, suggesting additional analysis or graphing is necessary.

Inflection Points

Inflection points occur where the concavity of a function changes, detected by solving for where the second derivative equals zero and examining sign changes around these points. Inflection points are the turning points in the graph between concave upwards and concave downwards segments.

To find potential inflection points, solve \( f''(x) = 12x^2 - 24x = 0 \). Factoring gives \( 12x(x - 2) = 0 \), suggesting potential inflection points at \( x = 0 \) and \( x = 2 \).

For verification, test intervals around these solutions. You need to check if \( f''(x) \) actually changes signs. If it changes from negative to positive or vice versa, you've confirmed an inflection point.

• For \( x = 0 \), testing reveals no change in sign of \( f''(x) \).
• For \( x = 2 \), \( f''(x) \) changes sign, confirming it as a valid inflection point.

Visually assessing the graph of the function can help confirm the nature of these changes and the overall behavior.