Problem 19 - 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. $$ g(x)=2 x^{2}-8 x, \quad[0,6] $$

Step-by-Step Solution

Answer

The absolute maximum of the function \(g(x)=2x^2-8x\) on the interval \([0,6]\) is \(0\) at \(x=0\) and the absolute minimum is \(-24\) at \(x=6\).

Step 1: Compute the derivative of the function

The derivative of \(g(x)=2x^2-8x\) is \(g'(x)=4x-8\).

Step 2: Find the critical points

You find these by setting the derivative equal to zero and solving for x. Thus, \(4x-8=0\) which gives the solution \(x=2\).

Step 3: Evaluate the function at critical points and endpoints

Evaluate \(g(x)\) at \(x=2\), \(x=0\), and \(x=6\). This gives the following: \(g(2)= -4\), \(g(0)=0\), and \(g(6)= -24\).

Step 4: Determine the absolute extrema

The maximum value of the function on the interval is \(0\) at \(x=0\) and the minimum value is \(-24\) at \(x=6\).

Key Concepts

Critical Points Calculus

In calculus, identifying critical points is crucial for understanding the behavior of a function, particularly when locating the peaks and valleys – the high and low points on a graph. Critical points are the points on a function where the derivative is either equal to zero or does not exist. To determine the critical points of a function, we look to its first derivative.

In the given exercise, the derivative of the function \( g(x) = 2x^2-8x \) is found to be \( g'(x) = 4x - 8 \). To find the critical points, we set the derivative equal to zero and solve for \( x \). This gives us \( 4x - 8 = 0 \), leading to a single critical point at \( x = 2 \). It's important to remember that not all critical points will lead to extrema; this is where further steps, like evaluating the function at these points and its endpoints or using the second derivative test, come into play.

Derivative of a Function

The derivative of a function represents the rate at which the function's output value changes with respect to changes in the input value. Essentially, it gives us the slope of the tangent line to the function's graph at any point. In the world of calculus, the derivative is a tool that enables us to predict and analyze the behavior of functions, especially when seeking to optimize real-world situations, such as maximizing profit or minimizing cost.

For our function \( g(x) = 2x^2-8x \), we derived the derivative function \( g'(x) = 4x - 8 \) by applying basic differentiation rules. After finding the derivative, we can then use it to identify critical points, as mentioned earlier, and to determine the increasing or decreasing intervals of the original function.

Absolute Extrema

The concept of absolute extrema refers to the highest (maximum) or lowest (minimum) points of a function over a given domain. When the domain is a closed interval, the Absolute Extrema Theorem states that a continuous function on that closed interval will indeed have both an absolute maximum and an absolute minimum. This is an important concept in various applications, such as physics and engineering, where we might need to find the greatest or smallest value that occurs in a particular situation.

To find the absolute extrema, we not only consider the critical points but also the endpoints of the interval. The function must be evaluated at these points to ascertain the largest and smallest values. In our problem, the function \( g(x) \) was evaluated at the critical point \( x = 2 \), and the endpoints of the interval \( x = 0 \) and \( x = 6 \). The function's absolute maximum was found to be \( 0 \) at \( x = 0 \), and the absolute minimum was \( -24 \) at \( x = 6 \), allowing us to conclude where the function reached its extremities within the closed interval.