Problem 73 - Calculus Single Variable

Question

Approximate the critical points and inflection points of the given function \(f\). Determine the behavior of \(f\) at each critical point. $$ f(x)=x /\left(1+e^{x}\right) $$

Step-by-Step Solution

Answer

Critical point at \( x = -0.567 \), function has a local maximum here. Inflection point at \( x = -0.852 \), where the function changes concavity.

Step 1: Find the First Derivative

To find the critical points, firstly, we need to find the first derivative of the function. The function is given as \( f(x) = \frac{x}{1+e^x} \). We'll apply the quotient rule \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \). Here, \( u = x \) and \( v = 1 + e^x \), so \( u' = 1 \) and \( v' = e^x \). The derivative is: \[ f'(x) = \frac{(1)(1 + e^x) - x(e^x)}{(1 + e^x)^2} \]. Simplify to get \[ f'(x) = \frac{1 + e^x - xe^x}{(1 + e^x)^2} \].

Step 2: Find Critical Points

Critical points occur where the first derivative is zero or undefined. Since \( f'(x) \) is never undefined, we set it to zero: \[ 1 + e^x - xe^x = 0 \]. Simplifying, we get \[ 1 = xe^x - e^x \]. Hence, find \( x \) such that \( x = \frac{1 + e^x}{e^x} \). Solving \( 1 = xe^x - e^x \) is complex analytically, so use numerical methods to approximate \( x \approx -0.567 \), which is our critical point.

Step 3: Determine the Behavior at Critical Points

Check the sign of \( f'(x) \) around the critical point \( x = -0.567 \). For \( x < -0.567 \), \( f'(x) > 0 \). For \( x > -0.567 \), \( f'(x) < 0 \). This means the function changes from increasing to decreasing at \( x = -0.567 \), indicating a local maximum.

Step 4: Find the Second Derivative

To identify inflection points, calculate the second derivative. Differentiating \[ f'(x) = \frac{1 + e^x - xe^x}{(1 + e^x)^2} \], apply the quotient rule to find \( f''(x) \). Skipping intermediate steps, the simplified second derivative is \[ f''(x) = \frac{-2e^x(1 + xe^x)}{(1 + e^x)^3} \].

Step 5: Find Inflection Points

Inflection points occur where the second derivative changes sign. Find \( x \) such that \( f''(x) = 0 \): \[ -2e^x(1 + xe^x) = 0 \]. Clearly, \( e^x eq 0 \), reducing this to \( 1 + xe^x = 0 \). Solving yields \( x \approx -0.852 \). Check sign change: \( f''(x) \) changes from positive to negative.

Step 6: Determine the Behavior at Inflection Points

Since \( f''(x) \) changes signs around \( x \approx -0.852 \), there is an inflection point here. The function changes concavity at this point.

Key Concepts

Critical Points

Critical points are locations on a graph where a function's slope is zero or undefined. They are pivotal because they indicate potential local minima, maxima, or saddle points, where the function may change its direction of increase or decrease. To find critical points for the function \( f(x) = \frac{x}{1+e^x} \), we calculate its first derivative \( f'(x) \). This involves differentiating using the quotient rule, resulting in:\[ f'(x) = \frac{1 + e^x - xe^x}{(1 + e^x)^2} \]To find critical points, we set \( f'(x) = 0 \), simplifying to \( 1 = xe^x - e^x \), and solve for \( x \). Through numerical approximation, we find a critical point at \( x \approx -0.567 \). At this point, analyzing the sign of \( f'(x) \) around \( x = -0.567 \) shows the function changes from increasing (when \( x < -0.567 \)) to decreasing (when \( x > -0.567 \)), thus indicating a local maximum.

Inflection Points

Inflection points occur where the graph of a function changes its concavity - from concave up to concave down or vice versa. These points are crucial for understanding the overall shape of the function's graph.Identifying inflection points requires the second derivative of the function. For \( f(x) = \frac{x}{1+e^x} \), we previously found the second derivative \( f''(x) \) using the quotient rule again:\[ f''(x) = \frac{-2e^x(1 + xe^x)}{(1 + e^x)^3} \]Inflection points are found by setting \( f''(x) = 0 \). Solving \( -2e^x(1 + xe^x) = 0 \) gives us \( x \approx -0.852 \) through approximation methods. At \( x = -0.852 \), \( f''(x) \) changes from positive to negative, indicating a change in concavity and confirming this as an inflection point.

Quotient Rule

The quotient rule is a method in calculus for finding the derivative of a function that is defined as a quotient of two functions. This rule is vital when working with functions that are divisions, such as our case with \( f(x) = \frac{x}{1+e^x} \).The rule is mathematically expressed as:\[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \]Where \( u \) and \( v \) are functions of \( x \). For our function, we set \( u = x \) and \( v = 1 + e^x \), giving us derivatives \( u' = 1 \) and \( v' = e^x \).Applying this rule enables us to find the first and second derivatives of the function efficiently, providing critical insights like critical points and inflection points.

Second Derivative

The second derivative of a function is the derivative of its first derivative. It offers insights into the function's concavity and helps in identifying inflection points.For the function \( f(x) = \frac{x}{1+e^x} \), the first derivative:\[ f'(x) = \frac{1 + e^x - xe^x}{(1 + e^x)^2} \]is differentiated once more using the quotient rule to obtain the second derivative:\[ f''(x) = \frac{-2e^x(1 + xe^x)}{(1 + e^x)^3} \]The second derivative is critical in determining where the function changes concavity, thus identifying inflection points. In our example, \( f''(x) = 0 \) indicated an inflection point at \( x \approx -0.852 \), where the function shifts from concave up to concave down.