Problem 19 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

In Exercises 17 through 20, sketch the graph of a function \(f\) that has all the given properties. a. \(f^{\prime}(x)>0\) when \(12\) c. \(f^{\prime \prime}(x)>0\) for \(x<2\) and for \(x>2\) d. \(f^{\prime}(1)=0\) and \(f^{\prime}(2)\) is undefined.

Step-by-Step Solution

Answer

The graph decreases, has a minimum at \(x=1\), increases between \(1 < x < 2\), and has a cusp at \(x=2\), then decreases again.

Step 1: Analyze the first derivative condition

Identify intervals where the first derivative, \(f'(x)\), is positive or negative. According to the problem, \(f'(x) > 0\) when \(1 < x < 2\), and \(f'(x) < 0\) when \(x < 1\) and \(x > 2\). This indicates \(f(x)\) is increasing on the interval \((1, 2)\) and decreasing on \((-\infty, 1)\) and \((2, \infty)\).

Step 2: Analyze the second derivative condition

Identify intervals where the second derivative, \(f''(x)\), is positive. According to the problem, \(f''(x) > 0\) for all \(x < 2\) and \(x > 2\). This indicates that \(f(x)\) is concave up on these intervals.

Step 3: Identify critical points

Given \(f'(x) = 0\) at \(x = 1\), there is a critical point at \(x = 1\). At \(x = 2\), \(f'(x)\) is undefined, indicating a possible cusp or vertical tangent.

Step 4: Sketch the graph

Using the information from the previous steps, sketch a graph that meets all the given conditions:1. The function decreases before \(x=1\).2. The function has a local minimum and starts to increase after \(x=1\).3. The function reaches \(x=2\), where it has a cusp or vertical tangent.4. After \(x=2\), it decreases again.

Key Concepts

First Derivative

The first derivative, denoted as \(f'(x)\), provides information about the rate of change of a function. It tells us whether the function is increasing or decreasing at a given point.

In the given exercise:

• \(f'(x) > 0\) when \(1 < x < 2\): This means the function is increasing in this interval.
• \(f'(x) < 0\) when \(x < 1\) and \(x > 2\): This indicates the function is decreasing in these intervals.

By analyzing \(f'(x)\), we learn how the function behaves over different intervals, helping us sketch its graph accordingly.

Second Derivative

The second derivative, denoted as \(f''(x)\), provides information about the concavity of the function. Concavity tells us whether the graph of a function curves upwards or downwards.

In this exercise:

• \(f''(x) > 0\) for \(x < 2\) and \(x > 2\): This means that the function is concave up in these intervals, forming a 'U' shape.

Understanding the concavity helps us refine our sketch by showing the curvature of the graph.

Critical Points

Critical points are where the first derivative is zero or undefined. These points are significant because they may indicate local maxima, minima, or points of inflection.

From the exercise:

• \(f'(x)=0\) at \(x=1\): This is a critical point, potentially indicating a local minimum or maximum.
• \(f'(x)\) is undefined at \(x=2\): This suggests a possible cusp or vertical tangent, a sharp turn in the graph.

Identifying critical points helps us locate significant changes in the function's behavior.

Concavity

Concavity describes how the graph bends. There are two types of concavity:

• Concave up (\(f''(x) > 0\)): Graph opens upwards like a 'U'.
• Concave down (\(f''(x) < 0\)): Graph opens downwards like an 'n'.

In our problem, since \(f''(x) > 0\) for all \(x < 2\) and \(x > 2\), the entire graph is concave up, which means it will look like multiple 'U' shapes within those intervals.

Local Minimum

A local minimum is a point where the function value is lower than all other nearby points.

In our given exercise:

• There is a local minimum at \(x=1\): The function decreases before \(x=1\) and increases after, creating a valley.

This is important as it provides a key feature in the graph sketch, showing where the function dips down the lowest within a certain region.