Problem 35 - University Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

To satisfy the conditions, use \(a=16\) for the local minimum at \(x=2\), and \(a=-1\) for the point of inflection at \(x=1\); both cannot be satisfied simultaneously.

Step 1: Understand the Requirements

The function is given as \(f(x) = x^2 + \frac{a}{x}\). We need to find the value of \(a\) such that: - The function has a local minimum at \(x=2\).- The function has a point of inflection at \(x=1\).

Step 2a: Find Conditions for Local Minimum at x=2

To have a local minimum at \(x=2\), the first derivative, \(f'(x)\), must be zero at \(x=2\) and the second derivative, \(f''(x)\), must be positive at \(x=2\).

Step 3a: Compute First Derivative and Set to Zero

Calculate the first derivative: \(f'(x) = 2x - \frac{a}{x^2}\). For the local minimum at \(x=2\), set \(f'(2) = 0\): \[2(2) - \frac{a}{2^2} = 0 \implies 4 = \frac{a}{4} \implies a = 16.\]

Step 4a: Compute Second Derivative and Verify Minimum

Calculate the second derivative: \(f''(x) = 2 + \frac{2a}{x^3}\). For a minimum at \(x=2\), ensure \(f''(2) > 0\): \[2 + \frac{2 \times 16}{2^3} = 2 + \frac{32}{8} = 6 > 0.\] This confirms a local minimum at \(x=2\).

Step 2b: Find Conditions for Point of Inflection at x=1

A point of inflection occurs where the second derivative changes sign, so \(f''(x)\) must be zero at \(x=1\).

Step 3b: Set Second Derivative to Zero at x=1

We already have \(f''(x) = 2 + \frac{2a}{x^3}\). Set \(f''(1) = 0\): \[2 + \frac{2a}{1^3} = 0 \implies 2 + 2a = 0 \implies 2a = -2 \implies a = -1.\]

Step 4b: Conclusion of Both Conditions

The conditions require different values: - For a local minimum at \(x=2\), \(a = 16\).- For a point of inflection at \(x=1\), \(a = -1\). The two conditions cannot be simultaneously satisfied by a single \(a\).

Key Concepts

Local Minimum

In calculus, a local minimum is a point where a function has a lower value than at any nearby points. This means there is a trough or dip at that point on the graph. To determine where a local minimum exists in a function, two main conditions are usually checked:

• The first derivative of the function at that point must be equal to zero, indicating a horizontal tangent line.
  
• The second derivative must be positive, which shows that the slope is increasing on either side of the point, confirming a dip.
  

In the exercise, for the function \(f(x) = x^2 + \frac{a}{x}\), a local minimum at \(x = 2\) requires us to set the first derivative to zero and verify that the second derivative is positive at this point.

Point of Inflection

A point of inflection on a graph is a spot where the curve changes the direction of its bend, moving from concave to convex or vice versa. This means the second derivative of the function changes its sign at this point.

• At a point of inflection, the second derivative is usually zero, but it's crucial that it changes sign around that point to confirm the inflection.
  
• In the case of our exercise, the point of inflection for the function \(f(x) = x^2 + \frac{a}{x}\) at \(x = 1\) requires us to set the second derivative to zero and ensure its sign changes around this value.

Finding a point of inflection can help understand how the graph behaves between peaks and troughs.

Second Derivative Test

The second derivative test is a helpful tool used to determine the nature of critical points found using the first derivative test. This test involves taking the second derivative of a function:

• If the second derivative at a critical point is positive, the function has a local minimum there.
  
• If the second derivative at a critical point is negative, the function has a local maximum.
  
• If the second derivative is zero, the test is inconclusive, meaning the point could be a minimum, maximum, or a point of inflection.
  

For the exercise, we used the second derivative \(f''(x) = 2 + \frac{2a}{x^3}\) to determine the condition for a local minimum at \(x = 2\) and to examine the inflection at \(x = 1\).

First Derivative Test

The first derivative test is used to classify critical points as either a local minimum or maximum or neither. It involves:

• Finding the first derivative of a function and setting it to zero to find critical points. These are points where the function's slope is zero, potentially indicating a peak or trough.
  
• By examining the sign of the derivative before and after these points, we can determine if the function is increasing or decreasing, revealing the nature of the critical point.
  
• If the derivative changes from positive to negative at a point, it's a local maximum. Conversely, a change from negative to positive indicates a local minimum.
  

In the exercise \(f'(x) = 2x - \frac{a}{x^2}\) was set to zero to identify critical points, validating whether \(a = 16\) produces a local minimum at \(x = 2\).