Problem 35 - Thomas Calculus

Question

What value of \(a\) makes \(f(x)=x^{2}+(a / x)\) have \begin{equation}\begin{array}{l}{\text { a. } \text { a local minimum at } x=2 ?} \\ {\text { b. a point of inflection at } x=1 ?}\end{array}\end{equation}

Step-by-Step Solution

Answer

a = 16 for a local minimum at x=2; a = -1 for a point of inflection at x=1.

Step 1: Understand Local Minimum Condition

A function has a local minimum at a point if the first derivative is zero and the second derivative is positive at that point. Start by finding the first derivative of the function.

Step 2: Differentiate the Function

The function is given by \(f(x) = x^2 + \frac{a}{x}\). Differentiate it to get \(f'(x) = 2x - \frac{a}{x^2}\). This first derivative will help us find conditions for local minima and other critical points.

Step 3: Set the First Derivative to Zero for Minimum at x=2

For a local minimum at \(x=2\), set \(f'(2) = 0\). Substitute \(x = 2\) into the derivative to get: \(2(2) - \frac{a}{2^2} = 0\). Solve this equation for \(a\).

Step 4: Solve for a in Minimum Condition

Simplify and solve the equation \(4 - \frac{a}{4} = 0\). This gives \(a = 16\). So, a local minimum occurs at \(x=2\) when \(a=16\).

Step 5: Understand Inflection Point Condition

A function has a point of inflection where the second derivative is zero and changes sign. Find the second derivative of the function to analyze inflection conditions.

Step 6: Differentiate Again for Second Derivative

Differentiate \(f'(x) = 2x - \frac{a}{x^2}\) to get the second derivative \(f''(x) = 2 + \frac{2a}{x^3}\).

Step 7: Set Second Derivative to Zero for Inflection at x=1

For a point of inflection at \(x=1\), set \(f''(1) = 0\). Substitute \(x=1\) into the second derivative: \(2 + \frac{2a}{1^3} = 0\). Solve this equation for \(a\).

Step 8: Solve for a in Inflection Condition

Simplify and solve \(2 + 2a = 0\) to find \(a = -1\). So, a point of inflection occurs at \(x=1\) when \(a=-1\).

Key Concepts

Local Minimum

In calculus, a local minimum occurs at a point where the function has a lower value than any nearby points. Mathematically, this means the **first derivative** at that point is zero, and the **second derivative** is positive.
For example, consider a hill and the valleys on its surface. The bottom of the valley, where the slope is zero, is the local minimum.

• First, find the first derivative of the function.
• Set the first derivative equal to zero and solve for the variable.
• Check that the second derivative is positive at this point.

In the exercise solution, applying these steps to the function \( f(x) = x^2 + \frac{a}{x} \), giving the first derivative \( f'(x) = 2x - \frac{a}{x^2} \).By setting \( f'(2) = 0 \), solving for \( a \), we find that at \( x = 2 \), the local minimum condition is satisfied with \( a = 16 \). This ensures the curve reaches a local minimum when \( a \) holds this value.

Inflection Point

An inflection point of a function is where the curve changes its concavity, such as shifting from concave up to concave down or vice versa. This occurs when the second derivative of the function is zero and changes sign at that point.
Think of an inflection point as the point on a roller coaster where after ascending, the path changes direction to descend, or vice versa.
To identify inflection points:

• Find the second derivative of the function.
• Set the second derivative equal to zero.
• Ensure the concavity changes around this point.

The exercise demonstrates this concept with \( f''(x) = 2 + \frac{2a}{x^3} \).Setting \( f''(1) = 0 \) leads to finding that \( a = -1 \) creates an inflection point at \( x = 1 \). This describes where and how the curve alters its shape.

Derivative

The derivative of a function provides insight into how the function changes with respect to its variable. It represents the slope of the tangent line at any given point on the curve.
Understanding the first derivative helps predict the function's behavior, including locating points of local minima or maxima and points where the function is increasing or decreasing.

• To find the derivative, apply differentiation rules such as the power rule and quotient rule.
• Use this information to analyze the function's slope.

In our function \( f(x) = x^2 + \frac{a}{x} \), the derivative \( f'(x) = 2x - \frac{a}{x^2} \) indicates the rate of change of the function. Thus, it acts as a tool to determine where the function has critical points like the local minimum at \( x = 2 \).

Second Derivative

The second derivative is the derivative of the first derivative, offering insight into the function's concavity and the behavior of its slope. While the first derivative tells you whether a function is increasing or decreasing, the second derivative reveals how the rate of increase or decrease is changing over time.
Think of the second derivative as the **acceleration** in constantly changing scenarios, indicating if the change itself is becoming slower or faster.
Here's how you can use the second derivative:

• Determine concavity: if the second derivative is positive, the function is concave up; negative, concave down.
• Identify inflection points where the second derivative changes sign.

In the exercise, \( f''(x) = 2 + \frac{2a}{x^3} \) becomes zero at \( x = 1 \) when \( a = -1 \), indicating a change in concavity and verifying the presence of an inflection point. Hence, understanding the second derivative is crucial for a complete picture of the function's characteristics.