Problem 17 - 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. \(f(x)=x^{4}-8 x^{2}+5\)

Step-by-Step Solution

Answer

Critical points: x = 0 (local max), x = ±2 (local min); Inflection points: x ≈ ±1.155.

Step 1: Find the First Derivative

To find the critical points of the function, we first need to find the first derivative of \(f(x)\). The function given is \(f(x) = x^4 - 8x^2 + 5\). The first derivative is computed as follows:\[ f'(x) = \frac{d}{dx}(x^4 - 8x^2 + 5) = 4x^3 - 16x. \]

Step 2: Set the First Derivative to Zero

To find critical points, set the first derivative equal to zero and solve for \(x\):\[ 4x^3 - 16x = 0. \]Factor out the common term:\[ 4x(x^2 - 4) = 0. \]This gives:\[ x = 0 \quad \text{or} \quad x^2 - 4 = 0. \]Solving \(x^2 - 4 = 0\):\[ x = \pm 2. \]Thus, the critical points are \(x = 0, \pm 2\).

Step 3: Verify the Nature of Critical Points

To determine if each critical point is a local maximum, minimum, or neither, examine the sign of \(f'(x)\) before and after each critical point. Alternatively, you can use the second derivative test:\[ f''(x) = \frac{d}{dx}(4x^3 - 16x) = 12x^2 - 16. \]Evaluate \(f''(x)\) at each critical point:- At \(x = 0\): \(f''(0) = 12(0)^2 - 16 = -16\), indicating a local maximum.- At \(x = 2\): \(f''(2) = 12(2)^2 - 16 = 32\), indicating a local minimum.- At \(x = -2\): \(f''(-2) = 12(-2)^2 - 16 = 32\), indicating a local minimum.

Step 4: Find the Second Derivative to Locate Inflection Points

Inflection points occur where the second derivative changes sign. We already calculated \(f''(x) = 12x^2 - 16\). Set this equal to zero and solve:\[ 12x^2 - 16 = 0 \]\[ 12x^2 = 16 \]\[ x^2 = \frac{4}{3} \]\[ x = \pm \sqrt{\frac{4}{3}} \approx \pm 1.155. \]Verify that \(f''(x)\) changes sign by testing intervals around these points.

Step 5: Interpretation Using a Graph

Using the graph, confirm the critical and inflection points. At \(x = 0\), the curve changes from concave up to concave down, confirming a local maximum. At \(x = \pm 2\), the curve changes from concave down to concave up, confirming local minima. Near \(x \approx \pm 1.155\), the curve changes concavity, confirming inflection points.

Key Concepts

First Derivative

To find critical points of a function, the first derivative plays a key role. It helps identify where the function's slope is zero or undefined. For our function, \( f(x) = x^4 - 8x^2 + 5 \), we find the first derivative as follows:\[ f'(x) = 4x^3 - 16x. \]Setting \( f'(x) \) to zero helps us discover the values of \( x \) where these slopes occur:\[ 4x(x^2 - 4) = 0. \]This reveals the critical points: \( x = 0 \) and \( x = \pm 2 \). Critical points are potential locations for local maxima, minima, or points of inflection, depending on the behavior of the function at these points.

Second Derivative

The second derivative of a function provides insight into the concavity of the graph. Concavity describes whether the graph curves upward or downward. Let's compute the second derivative for our function:\[ f''(x) = 12x^2 - 16. \]Examining \( f''(x) \) across different values of \( x \) allows us to determine changes in concavity. If the second derivative is positive, the graph is concave up; if negative, concave down. At critical points:

• \( f''(0) = -16 \) - concave down, indicating a local maximum.
• \( f''(2) = 32 \) and \( f''(-2) = 32 \) - concave up, indicating local minima.

Inflection Points

Inflection points are where the graph changes concavity, transitioning from concave up to concave down or vice versa. To locate these points, we set the second derivative to zero. For our function:\[ 12x^2 - 16 = 0. \]Solving this gives:\[ x = \pm \sqrt{\frac{4}{3}} \approx \pm 1.155. \]At these \( x \) values, test around these points shows a change in the sign of \( f''(x) \), confirming inflection points. This means at \( x \approx \pm 1.155 \), the graph's concavity changes.

Local Maximum

A local maximum occurs at a critical point where the graph's slope shifts from positive to negative. Using the second derivative test can help confirm this. In our scenario:- For \( x = 0 \), \( f''(0) = -16 \) is negative. The graph is concave down, establishing a peak, thus \( x = 0 \) is a local maximum.The shape of the curve near \( x = 0 \) will look like an upside-down "U" shape, indicating that this point is higher than nearby points.

Local Minimum

Local minima occur at critical points where the slope of the graph transitions from negative to positive. For this case, the second derivative test further verifies:- For both \( x = 2 \) and \( x = -2 \), \( f''(2) = 32 \) and \( f''(-2) = 32 \) are positive. Here, the graph is concave up, identifying \( x = 2 \) and \( x = -2 \) as local minima.Imagine the graph having a "U" shaped curve at these points, signifying they are valleys lower than the surrounding areas.