Problem 22 - Calculus with Concepts in Calculus

Question

Find all inflection points (if any) of the graph of the function. Then sketch the graph of the function. $$ f(x)=x^{3}+3 $$

Step-by-Step Solution

Answer

The inflection point is at \((0, 3)\).

Step 1: Find the Second Derivative

To find inflection points, we need the second derivative of the function. Start by finding the first derivative of the function \( f(x) = x^3 + 3 \). The first derivative is \( f'(x) = 3x^2 \). Now, differentiate \( f'(x) \) to find the second derivative, which is \( f''(x) = 6x \).

Step 2: Set the Second Derivative to Zero

Inflection points occur where the second derivative is zero or undefined and the concavity changes. Set the second derivative equal to zero: \( 6x = 0 \). Solve for \( x \): \( x = 0 \).

Step 3: Change of Concavity Check

To confirm it's an inflection point, check the sign of \( f''(x) \) around \( x = 0 \). For \( x < 0 \), choose \( x = -1 \), and \( f''(-1) = 6(-1) = -6 \) (negative, concave down). For \( x > 0 \), choose \( x = 1 \), and \( f''(1) = 6(1) = 6 \) (positive, concave up). Since the concavity changes from down to up at \( x = 0 \), \( x = 0 \) is indeed an inflection point.

Step 4: Find the Function Value at Inflection Point

Substitute \( x = 0 \) back into the original function to get the y-coordinate of the inflection point: \( f(0) = 0^3 + 3 = 3 \). Thus, the inflection point is \((0, 3)\).

Step 5: Sketch the Graph

The graph of \( f(x) = x^3 + 3 \) is a cubic function that is vertically shifted up by 3 units. The inflection point at \((0, 3)\) indicates a change from concave down to concave up. Sketch a curve starting lower left, rising through \((0, 3)\), and continuing upward to the upper right.

Key Concepts

Detecting Inflection Points Using the Second Derivative

Inflection points are fascinating features in the graph of a function where the curve changes its concavity. To find inflection points, the second derivative plays a crucial role. It involves these steps:

• First, find the first derivative of the function, which gives you a formula for the slope of the tangent line at any point.
• Differentiate the first derivative to get the second derivative. For example, if you have a function like \( f(x) = x^3 + 3 \), the first derivative is \( f'(x) = 3x^2 \) and the second derivative becomes \( f''(x) = 6x \).
• Set the second derivative equal to zero because inflection points occur where this derivative is zero or undefined.
• Solve this equation for \( x \) to find potential inflection points. In our example, \( 6x = 0 \) gives \( x = 0 \).
• Lastly, verify whether these points are indeed inflection points by testing the concavity change on either side, using values like \( x = -1 \) and \( x = 1 \).

Understanding Concavity and Its Role in Inflection Points

Concavity describes how a graph curves, and it's determined by the second derivative of a function. It can be broken down as follows:

• When the second derivative \( f''(x) \) is greater than zero, the graph is concave up. Imagine a bowl facing upwards.
• When \( f''(x) \) is less than zero, it's concave down, resembling an upside-down bowl.
• Inflection points are where the graph switches from one concavity to the other. For example, at \( x = 0 \) in the function \( f(x) = x^3 + 3 \), the concavity changes from downwards to upwards as confirmed by checking \( f''(-1) = -6 \) and \( f''(1) = 6 \).
• This change is crucial as it highlights a transition area in the graph's curvature pattern.

Exploring the Characteristics of a Cubic Function

Cubic functions are functions where the highest power of \( x \) is 3. They hold unique properties:

• The general form is \( f(x) = ax^3 + bx^2 + cx + d \). In our example, \( a = 1, b = 0, c = 0, \) and \( d = 3 \), making the function \( f(x) = x^3 + 3 \).
• Cubic functions can have one, two, or no real inflection points. Such points are identifiable where the second derivative is zero or changes signs.
• These functions often present an 'S' shaped curve, starting from the lower left, then crossing an inflection point, and moving towards the upper right. For the given function, when \( x = 0 \), the graph shifts from being concave down to concave up at the inflection point \((0, 3)\).
• Moreover, the constant term \( d \) affects the vertical shift of the whole graph, moving it up or down on the coordinate plane, which in this case is 3 units upwards.