Problem 24 - Calculus Single Variable

Question

Determine the intervals on which the given function \(f\) is concave up, the intervals on which \(f\) is concave down, and the points of inflection of \(f\). Find all critical points. Use the Second Derivative Test to identify the points \(x\) at which \(f(x)\) is a local minimum value and the points at which \(f(x)\) is a local maximum value. $$ f(x)=x-e^{x} $$

Step-by-Step Solution

Answer

The function is concave down everywhere; no points of inflection exist. Critical point at \(x = 0\) is a local maximum.

Step 1: Find the First Derivative

Start by finding the first derivative of the function. Given \(f(x) = x - e^x\), the first derivative is found using the derivative rules for polynomials and exponential functions: \(f'(x) = 1 - e^x\).

Step 2: Find the Second Derivative

To find the concavity of \(f(x)\), we need its second derivative. Differentiate \(f'(x) = 1 - e^x\) to get \(f''(x) = -e^x\).

Step 3: Determine Concavity

The function is concave up where \(f''(x) > 0\) and concave down where \(f''(x) < 0\). Since \(f''(x) = -e^x\) and the exponential function \(e^x\) is always positive, \(f''(x)\) is always negative, indicating \(f(x)\) is always concave down.

Step 4: Identify Points of Inflection

Points of inflection occur where \(f''(x) = 0\) or where the concavity changes. Since \(f''(x) = -e^x\) is never zero and does not change sign, there are no points of inflection.

Step 5: Find Critical Points

Critical points occur where \(f'(x) = 0\) or is undefined. Solving \(f'(x) = 1 - e^x = 0\) gives \(e^x = 1\), so \(x = 0\) is a critical point.

Step 6: Apply the Second Derivative Test

Use the second derivative test on the critical point \(x = 0\). Since \(f''(x) = -e^x < 0\) for all \(x\), including at \(x = 0\), \(f(x)\) has a local maximum at \(x = 0\).

Key Concepts

Critical Points

Critical points are values of \(x\) where the derivative of a function is either zero or undefined. These points are important because they can indicate where the function has a local maximum, local minimum, or a point of inflection.

For the function \(f(x) = x - e^x\), we calculate its first derivative as \(f'(x) = 1 - e^x\). By setting the first derivative equal to zero, \(1 - e^x = 0\), we solve for \(x\) and find that the function has a critical point at \(x = 0\).

It's essential to identify critical points because they tell us where to apply further tests like the second derivative test to determine the nature of these points.

Second Derivative Test

The second derivative test helps us determine whether a critical point is a local maximum, a local minimum, or neither. This test involves looking at the sign of the second derivative at the critical points.

For the function \(f(x) = x - e^x\), we found that the second derivative is \(f''(x) = -e^x\). Notice that \(f''(x)\) is always negative since exponentials \(e^x\) are always positive.

When applying the second derivative test at the critical point \(x = 0\), since \(f''(x) = -e^x < 0\), this indicates that \(f(x)\) has a local maximum at this point. In general, if \(f''(x) \gt 0\), it implies a local minimum, whereas if \(f''(x) \lt 0\), it indicates a local maximum at the critical point.

Concavity

Concavity describes how a function curves; it can be concave up (shaped like a cup) or concave down (shaped like a cap). The second derivative of a function tells us about its concavity.

For the function \(f(x) = x - e^x\), the second derivative is \(f''(x) = -e^x\). Since \(-e^x\) is negative for all real \(x\), the function is always concave down.

If \(f''(x) \gt 0\), the function is concave up, and if \(f''(x) \lt 0\), it is concave down. Understanding concavity can give insight into the function's behavior beyond just identifying maximums and minimums.

Points of Inflection

Points of inflection are points on the graph of a function where the concavity changes from up to down or vice versa. These points can be detected where the second derivative is zero or changes sign.

For our function \(f(x) = x - e^x\), the second derivative \(f''(x) = -e^x\) is always negative, indicating that concavity remains the same (concave down) across its domain. Thus, it cannot have any points of inflection because the concavity does not change sign.

Identifying points of inflection is crucial for providing a complete sketch of the graph, understanding its behavior, and pivotal in disciplines that necessitate precise curvature modeling.