Problem 41 - Single Variable Calculus

Question

In Exercises \(41-60\), find the absolute maximum and absolute minimum values, if any, of the function. $$ f(x)=x^{2}-x-2 \text { on }[0,2] $$

Step-by-Step Solution

Answer

The absolute maximum value of the function is 0 at \(x = 2\), and the absolute minimum value is \(-\frac{9}{4}\) at \(x = \frac{1}{2}\).

Step 1: Calculate the derivative of the function

First, we need to find the derivative of the function in order to identify critical points. The function is \(f(x) = x^2 - x - 2\). Let's find its derivative, \(f'(x)\): \[ f'(x) = \frac{d}{dx} (x^2 - x - 2)\] Using the power rule for differentiation, we get: \[ f'(x) = 2x - 1\]

Step 2: Find the critical points

Next, we need to find the critical points, i.e., the values of x for which the derivative is either 0 or undefined. In this case, the derivative is a linear function, so it is defined for all values of x. We only need to find the points where the derivative is equal to 0: \[ 2x - 1 = 0\] Solving for x, we get: \[ x = \frac{1}{2}\] So, there is one critical point, \(x = \frac{1}{2}\).

Step 3: Evaluate the function at critical points and endpoints

Now, we will evaluate the function at the critical point and the endpoints of the interval \([0, 2]\) to find the absolute maximum and minimum values of the function. The critical point is \(x = \frac{1}{2}\), and the endpoints are \(x = 0\) and \(x = 2\). Let's find the function values at these points: \[f(0)= 0^2 - 0 - 2 = -2\] \[f\left(\frac{1}{2}\right)= \left(\frac{1}{2}\right)^2 - \frac{1}{2} - 2 = -\frac{9}{4}\] \[f(2)= 2^2 - 2 - 2 = 0\]

Step 4: Determine the absolute maximum and minimum values

Now we will compare the values of the function at the critical point and endpoints: \[\begin{cases}f(0) = -2\\f\left(\frac{1}{2}\right) = -\frac{9}{4}\\f(2) = 0\end{cases}\] From these values, we can see that the absolute maximum value of the function is 0, which occurs at \(x=2\), and the absolute minimum value is \(-\frac{9}{4}\), which occurs at \(x=\frac{1}{2}\).

Key Concepts

Critical Points Calculus

Critical points in calculus are vital to understanding the behavior of differentiable functions. They are the points where a function's derivative is zero or undefined, which often indicates a local maximum, local minimum, or saddle point. For example, in the context of the function f(x) = x^2 - x - 2, we calculated its first derivative and set it equal to zero to find any critical points. The derivative f'(x) = 2x - 1 leads us to the critical point at x = 1/2. This is a potential location for an extrema on the function within the interval of consideration.

In studying calculus, it's essential to not only find these critical points but also to evaluate the function at these points to help determine the type of extremum each point represents. After finding the critical points, we typically use tests like the First or Second Derivative Test or evaluate their functional values, alongside the values at the boundaries of the domain (endpoints), to classify these extrema as absolute or relative.

Derivative of a Function

The derivative of a function represents the rate at which the function value changes with respect to a change in its input value, typically denoted as x. In more geometrical terms, it corresponds to the slope of the tangent line to the function's graph at a particular point. To find the extrema of a function, you first need to calculate its derivative. Using the power rule, one of the basic rules of differentiation, the derivative of f(x) = x^2 - x - 2 was found as f'(x) = 2x - 1.

The process of differentiation helps in locating the function's critical points, which are the candidates for the function's local and absolute extrema. Furthermore, by understanding the nature of a function's derivative, we can gain insight into the function's increasing or decreasing behaviors, and concavity, aiding in sketching the general shape of its graph.

Interval Evaluation

Interval evaluation is a critical step in determining the absolute extremal values of a function on a specific interval. After finding the critical points, we need to evaluate the function not only at these points but also at the endpoints of the given interval. This step ensures that we consider all possible locations for maximum and minimum values. In our exercise, f(x) = x^2 - x - 2 was evaluated at the critical point x = 1/2 and at the endpoints of the interval [0, 2].

By doing so, we determined that the absolute maximum and minimum of the function occur at these points within the closed interval. It's evident that absolute extrema can sometimes occur at the boundaries, which underscores the importance of evaluating the endpoints alongside the critical points. This comprehensive assessment of a function's value over an entire interval is a vital component of applied calculus problems, helping to understand the function's global behavior within a specific range.