Problem 50 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(H(x)\) have three continuous derivatives, and be such that \(H(1)=H^{\prime}(1)=H^{\prime \prime}(1)=0\), but \(H^{\prime \prime \prime}(1) \neq 0\). Does \(H(x)\) have a local maximum, local minimum, or a point of inflection at \(x=1\) ? Justify your answer.

Step-by-Step Solution

Answer

Point of inflection at \(x=1\).

Step 1: Understand the Problem

We need to determine the nature of the point at \(x = 1\) for the function \(H(x)\) given its derivatives at this point. Specifically, we need to find if it is a local maximum, local minimum, or a point of inflection.

Step 2: Recall the Derivative Test

The first and second derivatives of a function give information about its local maxima, minima, and points of inflection. The first derivative \(H'(x)\) indicates increasing or decreasing behavior, whereas the second derivative \(H''(x)\) indicates concavity.

Step 3: Analyze Given Data

We know \(H(1) = 0\), \(H'(1) = 0\), and \(H''(1) = 0\), but \(H'''(1) eq 0\). This implies that the rate of change of the second derivative at \(x = 1\) is not zero.

Step 4: Determine the Implications

Since \(H''(1) = 0\) and \(H'(1) = 0\), there's no indication of a local extremum. Also, \(H'''(1) eq 0\) indicates a change in concavity (the second derivative changes from concave upwards to concave downwards, or vice versa).

Step 5: Conclusion based on Calculus Theory

For \(H(x)\), because \(H''(1) = 0\) and \(H'''(1) eq 0\), according to the third derivative test, \(x = 1\) is a point of inflection. This is because the sign of \(H''(x)\) changes around \(x = 1\), due to the non-zero third derivative, indicating a change in concavity.

Key Concepts

Third Derivative Test

The third derivative test is a useful tool in calculus to determine the nature of a stationary point, especially when both the first and second derivatives are zero at that point. When a function's first derivative at a point is zero, it suggests that the function has a stationary point, which could either be a local extremum or a point of inflection. If the second derivative is also zero, the third derivative test comes into play.

In our scenario, we are considering the function \(H(x)\) at \(x = 1\) where \(H'(1) = 0\) and \(H''(1) = 0\). However, if \(H'''(1) eq 0\), the third derivative test tells us that the point is not a local extremum but a point of inflection. This is because a non-zero third derivative indicates a change in the concavity of the function as it passes through the point. Thus, for \(H(x)\), \(x = 1\) represents a point of inflection.

Concavity

Concavity describes the orientation of the curve of a function at a given point. It tells us whether a function is curving upwards like a bowl (concave up) or downwards like a frown (concave down). This is determined by the second derivative of the function.

When the second derivative \(H''(x)\) is positive, the function is concave up, and when it's negative, the function is concave down. At \(x = 1\) for the function \(H(x)\), \(H''(1) = 0\) initially provides no information about concavity, meaning it could be changing.

• A change from positive to negative (or vice versa) indicates a point of inflection.
• In conjunction with \(H'''(1) eq 0\), we confirm a change in concavity, cementing the presence of a point of inflection at that point.

In summary, analysis of concavity helps us understand the curvature and possible inflection points of a function.

Second Derivative

The second derivative, \(H''(x)\), provides key insights into the curvature of a function. It's crucial for determining both concavity and the existence of any points of inflection or local extrema. A non-zero second derivative informs us about the concavity direction:

• \(H''(x) > 0\): the function is concave up.
• \(H''(x) < 0\): the function is concave down.

When the second derivative equals zero, as it does at \(H''(1) = 0\), it can imply a potential point of inflection, especially if accompanied by \(H'''(1) eq 0\). Thus, in our specific problem, the zero value of the second derivative at \(x=1\) coupled with a non-zero third derivative confirms the point is a point of inflection.

Local Extremum

Local extrema occur at points where a function achieves a local maximum or minimum. The first and second derivatives are typically key to determining these. For a function \(H(x)\), if \(H'(x_0) = 0\) and \(H''(x_0) > 0\), \(x_0\) represents a local minimum; if \(H''(x_0) < 0\), it's a local maximum.

However, in the original problem, \(H(1) = 0\), \(H'(1) = 0\), and \(H''(1) = 0\), indicate a lack of local extrema at \(x = 1\). Since the second derivative test is inconclusive (because \(H''(1) = 0\)), and \(H'''(1) eq 0\) confirms a change in concavity, the point is identified as an inflection rather than a local extremum. This shows that derivatives play distinct roles in identifying the overall shape and critical points within the function.