Problem 22 - Calculus of a Single Variable

Question

Finding Extrema on a Closed Interval In Exercises \(17-36\) , find the absolute extrema of the function on the closed interval. $$ f(x)=2 x^{3}-6 x,[0,3] $$

Step-by-Step Solution

Answer

The maximum value of the function on the interval [0,3] is 9 and it occurs at \(x = 3\), while the minimum value is -4 and it's found at \(x = 1\).

Step 1: Calculation of Derivative

First step is to compute the derivative of the given function. The derivative of \(f(x) = 2x^3 - 6x\) is \(f'(x) = 6x^2 - 6\).

Step 2: Finding the Critical Points

Next, we need to set the derivative equal to zero to find the critical points, i.e., solve the equation \(6x^2 - 6 = 0\). This simplifies to \(x^2 = 1\) and the solutions of this equation are \(x = 1\) and \(x = -1\). However, since we are defining \(x\) to be in the interval [0,3], only \(x = 1\) is relevant.

Step 3: Determine the Function Values at the Critical Points and Ends of the Interval

The critical point and the two endpoints will be evaluated in the function, which will result in three values: \(f(0) = 2(0)^3-6(0) = 0\), \(f(1) = 2(1)^3-6(1) = -4\), and \(f(3) = 2(3)^3 - 6(3) = 9\)

Step 4: Identify the Maximum and Minimum Values

The maximum value is the greatest one of these, which is 9, and it is achieved at x=3. The minimum value is the smallest one of these, which is -4 and is obtained at x=1.

Key Concepts

Derivatives

Derivatives play a crucial role when identifying extrema of a function. They help determine how a function's graph behaves. To find the derivative of a function like our example, we use basic differentiation rules. The original function given is \( f(x) = 2x^3 - 6x \). By applying the power rule:

• The derivative of \( x^n \) is \( nx^{n-1} \).
• Hence, the derivative of \( 2x^3 \) is \( 6x^2 \).
• The derivative of \( -6x \) is \( -6 \).

Combining these, we find the derivative \( f'(x) = 6x^2 - 6 \).
This expression shows us how the slope of the tangent line to the curve changes as \( x \) changes.

Closed Interval

A closed interval, such as \([0, 3]\), includes both endpoints. It's written as \( x \) values where \( x \) is between 0 and 3, inclusive. In the context of finding absolute extrema, the closed interval is essential since it tells us where to check for maximum and minimum values.
For functions on closed intervals, extrema can occur at:

• Critical points inside the interval.
• Endpoints of the interval.

Here, we check all these potential locations to ensure we find the absolute maximum and minimum values within \([0, 3]\).

Critical Points

Critical points are places where the derivative of the function is zero or undefined, and they indicate potential locations for extrema.
For the function \( f(x) = 2x^3 - 6x \), we have the derivative \( f'(x) = 6x^2 - 6 \). To find critical points, set the derivative equal to zero:
\( 6x^2 - 6 = 0 \).
Simplifying this gives \( x^2 = 1 \), leading to \( x = 1 \) and \( x = -1 \). However, since the domain is restricted to \([0, 3]\), only \( x = 1 \) is relevant.
This critical point is then used to evaluate the function's behavior on the specified interval.

Function Evaluation

Function evaluation involves calculating the function's value at specific points. For determining absolute extrema, evaluate the function at critical points and at the boundaries of the interval.
For example, with the function \( f(x) = 2x^3 - 6x \), calculate:

• At \( x = 0 \), \( f(0) = 0 \).
• At \( x = 1 \), \( f(1) = -4 \).
• At \( x = 3 \), \( f(3) = 18 \).

By evaluating at these points, we can determine the absolute maximum and minimum values within the interval. Here, the maximum is 18 at \( x = 3 \) and the minimum is -4 at \( x = 1 \).