Problem 19 - Calculus for Biology and Medicine

Question

Determine all inflection points. $$ f(x)=x^{3}-2, x \in \mathbf{R} $$

Step-by-Step Solution

Answer

The function has an inflection point at \( x = 0 \).

Step 1: Find the Second Derivative

To find the inflection points, we first need the second derivative of the function. Start by finding the first derivative of \( f(x) = x^3 - 2 \), which is \( f'(x) = 3x^2 \). Then, find the second derivative: \( f''(x) = 6x \).

Step 2: Set the Second Derivative to Zero

Inflection points occur where the second derivative changes signs, which means we need to find where \( f''(x) = 0 \). Setting the second derivative equal to zero, we solve the equation: \( 6x = 0 \).

Step 3: Solve for x

Solve the equation \( 6x = 0 \) for \( x \). Dividing both sides by 6, we find \( x = 0 \).

Step 4: Verify the Inflection Point

To confirm \( x = 0 \) is an inflection point, check if the sign of \( f''(x) \) changes around \( x = 0 \). When \( x < 0 \), \( f''(x) = 6x < 0 \). When \( x > 0 \), \( f''(x) = 6x > 0 \). Since the sign changes, \( x = 0 \) is indeed an inflection point.

Key Concepts

Understanding the Second Derivative

The second derivative of a function provides us with crucial information about the concavity of the graph. For the function given, the original exercise tells us to find the second derivative, which we start by taking the first derivative of the function. The first derivative, \( f'(x) = 3x^2 \), reveals information about the slopes of the function but does not suffice to pinpoint changes in concavity. By taking the second derivative, \( f''(x) = 6x \), we can examine how the rate of change of the slope behaves.

The second derivative gives us:

• The concavity of the function: If \( f''(x) > 0 \), the graph is concave up; if \( f''(x) < 0 \), it is concave down.
• Potential inflection points: These occur where \( f''(x) \) might change signs.

Thus, analyzing \( f''(x) \) not only lets us determine the behavior of the function but also helps identify points where the graph shifts from curving upwards to curving downwards or vice versa.

The Importance of Sign Change

Sign changes in the second derivative are pivotal to locating inflection points. Specifically, an inflection point arises where the second derivative changes its sign. This sign change indicates a transition in concavity from upwards to downwards or vice versa—not just a flattening of the curve.
To explore, set \( f''(x) = 6x \) equal to zero and solve for \( x \). Doing so gives \( x = 0 \), but this is just a candidate for an inflection point. To confirm it's truly an inflection point, we must assess whether \( f''(x) \) changes from positive to negative or negative to positive across \( x = 0 \):

• For \( x < 0 \), \( f''(x) = 6x < 0 \) indicating the curve is concave down.
• For \( x > 0 \), \( f''(x) = 6x > 0 \) showing the curve is concave up.

Because there is a change in sign, we can confirm that \( x = 0 \) is indeed an inflection point. Recognizing sign change in derivatives is critical for accurately identifying transitions in function behavior.

Tools in Calculus Problem Solving

When solving calculus problems, finding inflection points involves several key skills:
1. **Derivative Operations**: Grasping both first and second derivatives is essential. Each derivative provides different insights—first derivatives for slope and second derivatives for concavity. In our exercise, finding \( f'(x) = 3x^2 \) and \( f''(x) = 6x \) equips us with a complete view of how the original function behaves.
2. **Setting Derivatives to Zero**: Doing this reveals critical points. For inflection points, we set \( f''(x) = 0 \) to find candidates.
3. **Sign Analysis**: After obtaining potential points from the derivative, analyzing sign before and after these values confirms whether these points are true inflection points. As we did with \( x=0 \), ensuring these confirmatory steps distinguishes inflection points from mere critical points.
Using these steps effectively allows for a detailed understanding of the function's geometry, specifically its curves and direction. Mastery of these tools is essential for navigating more complex calculus problems that involve understanding the deeper characteristics of functions.