Problem 23 - Applied 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

(a) Impossible, (b) Possible, (c) Possible.

Step 1: Understanding the Information

We know that the function has a specific value and derivatives at a certain point: \(f(5)=20\), \(f'(5)=2\), and \(f''(x)<0\) for \(x \geq 5\). This means the function is concave down in the given interval, which implies the growth rate is decreasing.

Step 2: Applying the First Derivative

The derivative \(f'(5) = 2\) indicates that at \(x=5\), the function increases at a rate of 2 units per unit increase in \(x\). Given \(f''(x)<0\), this rate of increase decreases as \(x\) increases. Thus, at \(x = 7\), the increase rate is less than 2.

Step 3: Estimating Possible Function Values

To estimate \(f(7)\), consider the maximum scenario using constant rate 2 from the derivative. Over the interval from \(x=5\) to \(x=7\), we'd expect the function to increase by about \((2 \, \text{units} / \text{unit in } x) \times 2 \, \text{units} = 4\) units, as the rate is slowing. Thus, \(f(7)\) can be at most \(20 + 4 = 24\).

Step 4: Evaluating Each Option

- Option (a) 26: Impossible, as it's larger than 24, which is the upper bound for \(f(7)\). - Option (b) 24: Possible, as it fits the estimated maximum of how much \(f(x)\) can grow.- Option (c) 22: Also possible, as \(f(x)\) could grow by less than 4 units.

Key Concepts

Understanding Derivatives

In calculus, derivatives are a fundamental tool to understand how a function behaves. They tell us the rate at which a function is changing at any given point. For example, if you have a function representing the distance a car travels over time, its derivative would represent the car's speed at each moment.

In this exercise, we have the first derivative, \(f'(x)\), which shows how the function \(f(x)\) changes as \(x\) increases. At \(x = 5\), \(f'(5) = 2\), meaning the function's rate of change is 2 units per unit change in \(x\). This tells us that at exactly \(x = 5\), the function is growing. However, since the second derivative is negative, this rate of grow slows down as \(x\) increases.

Exploring Concavity

Concavity is about the curvature of a function's graph. It describes how a function bends or curves. A function is concave up if it forms a cup-like shape, meaning it bends upwards, and it's concave down if it forms a cap-like shape, bending downwards.

In our specific problem, we have information indicating that the second derivative, \(f''(x)\), is less than zero for \(x \geq 5\). This negative second derivative implies the function is concave down from \(x = 5\) onwards.

• Concave down means the slope of the tangent line is decreasing.
• Even if the function still increases, it does so at a reducing pace.

Thus, as \(x\) moves from 5 to 7, \(f'(x)\) decreases which indicates that the function's growth is slowing down because of discarding a steeper slope.

Analyzing Function Growth

Understanding how a function grows is crucial in predicting its behavior in different intervals. In this exercise, we're tasked with finding how much the function value could increase between \(x = 5\) and \(x = 7\).
Initially, considering the first derivative \(f'(5) = 2\), if the derivative were constant (which it is not), the function value may increase by 4 units over the interval. So we might expect \(f(7)\) to be \(24\), starting from \(f(5) = 20\). However, due to the concavity, the actual increase is less than 4.

Let's apply this to the options:

• For \(f(7) = 26\), this result would signify an increase of 6 units, which is impossible due to the decreasing rate of growth.
• If \(f(7) = 24\), this is the maximum possible increase, making it a feasible outcome.
• For \(f(7) = 22\), this reflects a lesser increase than the maximum possible, allowing it to be a possible solution.

In summary, analyzing how the concavity impacts function growth gives insight into how values may change over intervals and helps determine possible outcomes.