Problem 13 - Essential Calculus Early Transcendentals

Question

Suppose \(f^{\prime \prime}\) is continuous on \((-\infty, \infty)\) (a) If \(f^{\prime}(2)=0\) and \(f^{\prime \prime}(2)=-5,\) what can you say about \(f ?\) (b) If \(f^{\prime}(6)=0\) and \(f^{\prime \prime}(6)=0,\) what can you say about \(f ?\)

Step-by-Step Solution

Answer

(a) Local maximum at \( x = 2 \). (b) No sufficient information at \( x = 6 \).

Step 1: Identify Critical Point for Part (a)

Since \( f'(2) = 0 \), \( x = 2 \) is a critical point of \( f \). This means that \( f \) has a stationary point at \( x = 2 \), where the derivative changes sign or remains zero.

Step 2: Analyze Second Derivative Test for Part (a)

Since \( f''(2) = -5 \), which is less than zero, the function \( f \) has a local maximum at \( x = 2 \). The negative value of the second derivative indicates that the slope of \( f' \) comes downward at the critical point, confirming a local maximum.

Step 3: Identify Critical Point for Part (b)

Given that \( f'(6) = 0 \), \( x = 6 \) is also a critical point of \( f \) where the derivative is zero, indicating a stationary point like that in part (a).

Step 4: Analyze Second Derivative for Part (b)

For \( x = 6 \), \( f''(6) = 0 \) does not give sufficient information to determine whether there is a maximum, minimum, or neither at \( x = 6 \). Further investigation such as higher derivatives or context is needed to draw any conclusion.

Key Concepts

Critical Points

Understanding critical points helps us analyze the behavior of a function. A critical point occurs where the derivative of the function, denoted as \( f'(x) \), is equal to zero. This indicates that the slope of the tangent to the curve is horizontal at this point.

• This could mean that the function reaches a peak, a valley, or experiences a change in direction.
• In our example, \( x = 2 \) and \( x = 6 \) are critical points since the derivative is zero at these points.

Second Derivative Test

The second derivative test is a valuable tool to determine the nature of a critical point. Calculating \( f''(x) \), the second derivative, helps us ascertain whether the critical point is associated with a local maximum, minimum, or neither.

• If \( f''(x) > 0 \), the function has a local minimum at that point.
• If \( f''(x) < 0 \), there is a local maximum.
• If \( f''(x) = 0 \), the test is inconclusive, and we might need further analysis.

In part (a) of the example, with \( f''(2) = -5 \), the test confirms a local maximum.

Stationary Points

A stationary point is where the first derivative \( f'(x) \) is zero, indicating no change in the slope of the function, either upwards or downwards.

• This is crucial as it signals a possible maximum, minimum, or inflection point.
• Stationary points occur at \( x = 2 \) and \( x = 6 \) in our problem.

It's essential to perform second derivative or further analyses to confirm the nature of these points.

Local Maximum

A local maximum occurs at a critical point where the function reaches its highest value in a small surrounding region.

• The slope of the graph changes from positive to negative, confirmed by a negative second derivative.

Using our example, since \( f''(2) = -5 \) is less than zero, it indicates a local maximum at \( x = 2 \), implying that \( f(x) \) peaks around this point, before beginning to decrease.

Higher Derivatives

When the second derivative test is inconclusive, we may need to evaluate higher derivatives to determine the behavior of the function at critical points.

• The third derivative or additional derivatives could provide more clues about concavity.
• This approach can be particularly useful in cases where \( f''(x) = 0 \), as seen in part (b) for \( x=6 \).

In such situations, examining these higher-order derivatives helps conclude whether the point corresponds to an inflection point or continues as a flat region.