Problem 41 - Calculus

Question

A function \(f\) has \(f(5)=20, f^{\prime}(5)=2,\) and \(f^{\prime \prime}(x)<0\) for \(x \geq 5 .\) Which of the following are possible values for \(f(7)\) and which are impossible? (a) 26 (b) 24 (c) 22

Step-by-Step Solution

Answer

Possible: 22; Impossible: 24, 26.

Step 1: Understand the Problem

We are given a function \(f\) with specific conditions: \(f(5) = 20\), \(f'(5) = 2\), and \(f''(x) < 0\) for \(x \geq 5\). We need to determine which of the given values (26, 24, 22) are possible for \(f(7)\). The condition \(f''(x) < 0\) implies the function is concave downwards.

Step 2: Analyze the Derivative and Concavity

Since \(f''(x) < 0\), this means \(f'(x)\) is decreasing for \(x \geq 5\). At \(x = 5\), \(f'(5) = 2\). As \(x\) increases, \(f'(x)\) decreases. Therefore, between \(x = 5\) and \(x = 7\), \(f'(x)\) will be less than 2, moving towards 0.

Step 3: Estimate the Increment Over the Interval [5, 7]

From \(x = 5\) to \(x = 7\), let's consider the area under the derivative graph (which is a decreasing linear function because the second derivative is negative). It's initially 2 at \(x = 5\), decreasing to some value less than 2 as \(x\) approaches 7. Therefore, the increase in \(f(x)\) from 5 to 7 is less than \(2 \times 2 = 4\) because the slope is decreasing.

Step 4: Apply the Increment to Determine the Range of Possible Values

Since the increase from \(x = 5\) to \(x = 7\) must be less than 4, starting from \(f(5) = 20\), \(f(7)\) must be less than 24. Therefore, \(f(7) = 26\) is impossible, \(f(7) = 24\) is also impossible, and \(f(7) = 22\) is possible because 22 is within the feasible range given the constraints.

Key Concepts

Derivative

The derivative of a function is a powerful concept in calculus that helps us understand how a function behaves. It's often referred to as the "slope" of the function at any given point. By looking at the first derivative, denoted as \(f'(x)\), you can determine the rate at which the function \(f(x)\) is changing at any point along the curve.

• If \(f'(x) > 0\), the function is increasing at that point.
• If \(f'(x) < 0\), the function is decreasing.
• If \(f'(x) = 0\), the function might be at a local maximum or minimum or a point of inflection.

For the original exercise, \(f'(5) = 2\), suggests that around \(x=5\), the function \(f\) is increasing. However, as you move away from \(x=5\), since it decreases (due to \(f''(x)<0\)), the increase becomes smaller.

Function Behavior

Understanding the behavior of functions is crucial, especially when determining values over specific intervals. A function's behavior is influenced by its first and second derivatives.

• The first derivative, \(f'(x)\), tells us whether the function is increasing or decreasing.
• The second derivative, \(f''(x)\), indicates the concavity of the function, helping us discern the shape of its graph.

For a function where \(f''(x) < 0\), as in our exercise, the function is concave downwards. This means it bends down like a sad face, linking to the notion that \(f'(x)\) is decreasing.
In the case of \(x \geq 5\), the downward concavity ensures that although \(f(x)\) might still increase due to \(f'(5) = 2\), the rate of increase slows down continuously.

Increasing and Decreasing Functions

Functions exhibit different behaviors, depending on whether they are increasing or decreasing. Determining these characteristics requires examining the sign of the first derivative \(f'(x)\).

• If \(f'(x) > 0\) for a range of \(x\), the function is increasing over that range.
• If \(f'(x) < 0\), the function is decreasing.
• If \(f''(x) < 0\), it adds an overlay of concavity, meaning even though \(f'(x)\) might be initially positive, it's decreasing overall.

In this exercise, from \(x = 5\) to \(x = 7\), \(f(x)\) is not just a simple increasing function. Because \(f''(x) < 0\), any increase observed as "less steep," meaning \(f(x)\) will not climb as high as it would if the function were concave up.
This decreasing concavity, despite \(f'(5)\) starting at 2, influences our prediction of possible function values such as 26 and 24 being too high, leaving 22 as a possible realistic value for \(f(7)\).