Finding Points of Inflection in a Function: A Step-by-Step Guide with Example

Points of Inflection

Points of inflection occur in a function when the concavity changes

Points of inflection occur in a function when the concavity changes. In other words, at a point of inflection, the function changes from being concave up to being concave down, or vice versa. These points are important in calculus because they indicate a change in the curvature of the graph.

To determine the points of inflection, we need to follow a few steps:

1. Take the second derivative of the function: The second derivative gives us information about the concavity of the function. For example, if the second derivative is positive, it means the function is concave up, and if it is negative, the function is concave down.

2. Find the critical points: Set the second derivative equal to zero and solve for x. These values of x represent potential points of inflection.

3. Determine the concavity intervals: To determine if the potential points of inflection are actual points of inflection, we need to analyze the concavity of the function in each interval between the critical points.

4. Test each interval: Choose a value within each interval and plug it into the second derivative. If the result is positive, the function is concave up in that interval. If the result is negative, the function is concave down.

5. Identify the points of inflection: If the concavity changes from positive to negative or from negative to positive at a potential point of inflection, then that point is a valid point of inflection.

It’s important to note that not all potential points of inflection are actual points of inflection. Some potential points may not satisfy the concavity change condition.

Let’s go through an example to illustrate these steps:

Find the points of inflection of the function f(x) = x^3 – 3x^2 + 2x.

1. Take the second derivative: The first derivative of f(x) is f'(x) = 3x^2 – 6x + 2. Differentiating again, we get the second derivative f”(x) = 6x – 6.

2. Find the critical points: Setting f”(x) = 0, we have 6x – 6 = 0. Solving for x, we find x = 1.

3. Determine the concavity intervals: We can see that the critical point x = 1 is the only one, so we have one interval (-∞, 1) and one interval (1, ∞).

4. Test each interval: Let’s choose x = 0 for the interval (-∞, 1) and x = 2 for the interval (1, ∞). Plugging these values into f”(x), we get f”(0) = -6 and f”(2) = 6.

– In the interval (-∞, 1), f”(0) = -6, so the function is concave down.
– In the interval (1, ∞), f”(2) = 6, so the function is concave up.

5. Identify the points of inflection: In this case, the concavity changes from concave down to concave up at x = 1. Therefore, the point (1, f(1)) is a point of inflection.

In summary, the point (1, f(1)) is the only point of inflection for the function f(x) = x^3 – 3x^2 + 2x.

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