Ch. 4 - Applications of Derivatives

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 4 - Applications of Derivatives

Problem 4.5.38b

Chapter 4, Problem 4.5.38b

38. What values of a and b make f(x) = x^3 + ax^2 + bx have
b. a local minimum at x = 4 and a point of inflection at x = 1?

Verified step by step guidance

To find the values of a and b, we need to use the conditions given: a local minimum at x = 4 and a point of inflection at x = 1.

First, find the first derivative of f(x): f'(x) = 3x^2 + 2ax + b. For a local minimum at x = 4, f'(4) must be 0.

Substitute x = 4 into f'(x) to get the equation: 3(4)^2 + 2a(4) + b = 0. Simplify this to find one equation in terms of a and b.

Next, find the second derivative of f(x): f''(x) = 6x + 2a. For a point of inflection at x = 1, f''(1) must be 0.

Substitute x = 1 into f''(x) to get the equation: 6(1) + 2a = 0. Solve this equation to find the value of a, and then use it in the first equation to find b.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Local Minimum

A local minimum of a function occurs at a point where the function value is lower than at nearby points. To find a local minimum, the first derivative of the function is set to zero, indicating a critical point, and the second derivative is checked to be positive, confirming a local minimum. In this problem, the function f(x) must have a local minimum at x = 4.

Point of Inflection

A point of inflection is where the function's concavity changes, which can be identified by the second derivative. At a point of inflection, the second derivative is zero, but the sign of the second derivative changes around this point. For the function f(x), there must be a point of inflection at x = 1, meaning the second derivative changes sign at this point.

Derivatives

Derivatives are fundamental in calculus for analyzing the behavior of functions. The first derivative provides information about the slope and critical points, while the second derivative gives insights into concavity and points of inflection. Solving the problem requires calculating both the first and second derivatives of f(x) = x^3 + ax^2 + bx to find values of a and b that satisfy the given conditions.

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