Problem 3 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

In Problems 1 through 12: (a) Find all critical points. (b) Find \(f^{\prime \prime} .\) Use the second derivative, wherever possible, to determine which critical points are local maxima and which are local minima. If the second derivative test fails or is inapplicable, explain why and use an alternative method for classifying the critical point. $$ f(x)=x^{3}+\frac{9}{2} x^{2}-12 x+\frac{3}{2} $$

Step-by-Step Solution

Answer

The critical points are -4 and 1. -4 is a local maximum and 1 is a local minimum.

Step 1: Find the first derivative

Differentiate the function \(f(x)=x^{3}+\frac{9}{2} x^{2}-12 x+\frac{3}{2}\) to obtain the first derivative. Using the power rule, \(f'(x) = 3x^{2} + 9x - 12\).

Step 2: Identify the critical points

Set the first derivative equal to zero and solve for x to find the critical points of the function. The critical points are values of x that satisfy the equation \(3x^{2} + 9x - 12 = 0\). Solving this, we get \(x = -4,1\).

Step 3: Find the second derivative

Differentiate \(f'(x) = 3x^{2} + 9x - 12\) to obtain the second derivative. Using the power rule, \(f''(x) = 6x + 9\).

Step 4: Apply the second derivative test

Use the second derivative to classify the critical points. Substitute the critical points calculated in Step 2 into \(f''(x) = 6x + 9\). If the result is positive, then the critical point is a local minimum, if it's negative, it's a local maximum, and if the result is zero, the test fails. Upon substituting, \(f''(-4) = -15\) and \(f''(1) = 15\), so -4 is a local maximum and 1 is a local minimum.

Key Concepts

Second Derivative Test

The second derivative test is a method used to determine whether a critical point of a function is a local maximum, local minimum, or a point of inflection. It involves analyzing the second derivative of the function, denoted as \(f''(x)\), at the critical points.
To apply the second derivative test:

• First, find the critical points by setting the first derivative, \(f'(x)\), to zero.
• Then, calculate the second derivative \(f''(x)\).
• Evaluate \(f''(x)\) at each critical point.
• If \(f''(x) > 0\) at the critical point, the function has a local minimum there.
• If \(f''(x) < 0\) at the critical point, the function has a local maximum.
• If \(f''(x) = 0\), the test is inconclusive, and further analysis is needed to classify the point.

In our example, the second derivative test helps distinguish between a local maximum at \(x = -4\) and a local minimum at \(x = 1\).

Local Maxima

A local maximum of a function is a point where the function value is greater than the values at all nearby points. In mathematical terms, if \(f(x)\) has a local maximum at \(x = c\), then \(f(c) \ge f(x)\) for all \(x\) in the vicinity of \(c\).
To determine if a critical point is a local maximum:

• Check the sign of the second derivative at the critical point. If \(f''(x) < 0\), it's a local maximum.
• If the second derivative is zero, examine the nature of \(f'(x)\) around the point or use other methods like the first derivative test.

From the exercise, using the second derivative test, \(x = -4\) is concluded to be a local maximum as \(f''(-4) = -15\) which is less than zero.

Local Minima

A local minimum is a point on the function where the function value is less than that of any nearby points. At \(x = c\), if \(f(c) \le f(x)\) for all \(x\) near \(c\), then \(c\) is a local minimum.
Identifying a local minimum involves:

• Using the second derivative test; if \(f''(x) > 0\), the point is a local minimum.
• If the second derivative is zero, further investigation like the first derivative test might be needed.

In our example, \(x = 1\) is identified as a local minimum since \(f''(1) = 15\) which is greater than zero.

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the derivative of a function. The derivative represents the rate at which the function value changes as its input changes. It is often used to identify critical points, optimize functions, and understand function behavior.
In the given exercise:

• The first derivative \(f'(x)\) was derived from \(f(x)\) to identify critical points by setting it to zero.
• The second derivative \(f''(x)\) was then calculated to classify these critical points using the second derivative test.

By effectively using differentiation, we can better understand the nature of functions and solve various mathematical problems.

Power Rule

The power rule is a basic rule in differentiation that makes it easy to find the derivative of polynomial terms. It states that if \(f(x) = x^n\), then the derivative \(f'(x) = n \cdot x^{n-1}\).
This rule simplifies the process of calculating derivatives for functions involving powers of \(x\).
In the exercise, we applied the power rule repeatedly:

• First, to find \(f'(x) = 3x^2 + 9x - 12\) from \(f(x) = x^3 + \frac{9}{2}x^2 - 12x + \frac{3}{2}\).
• Then, again to find \(f''(x) = 6x + 9\) from \(f'(x) = 3x^2 + 9x - 12\).

This rule is essential for handling polynomial functions, making it a crucial tool for calculus students.