Question

In Exercises \(1-6,\) use the First Derivative Test to determine the local extreme values of the function, and identify any absolute extrema.Support your answers graphically. \(y=x^{2}-x-1\)

Short answer

By using the First Derivative Test, a local minimum at \(x=1/2\) is found, and this point is also the absolute minimum.

Step-by-step solution

Differentiate the function

To start, the first derivative of the function y=x^{2}-x-1 is calculated by applying the power rule, giving \(y'=2x-1\).

Find the critical points

The critical points occur where the derivative equals zero or is undefined. As the derivative \(y'=2x-1\) is a linear function, it is defined for all x. The critical points are found by solving the equation \(y'=0\). Solving \(2x-1=0\) yields \(x=1/2\).

Applying the First Derivative Test

To classify the critical points, pick a number on either side of \(x = 1/2\) and evaluate the sign of the derivative at these points. If, for example, \(x=0\) is selected, \(y'(-0)= -1\), showing the function is decreasing. If \(x=1\) is picked, \(y'(1)=1\), which tells us the function is increasing. Therefore, there is a local minimum at \(x=1/2\).

Find any absolute extrema

As the function is a polynomial, it is defined for all real numbers. Therefore, the absolute extrema could occur at \(x=1/2\), or as \(x\) tends toward infinity or negative infinity. Evaluating at these points gives \(y(1/2)=-3/4\) and as \(x\) grows large or small, \(y\) also grows large, showing that there is an absolute minimum at \(x=1/2\).

Key concepts

Critical Points

In the realm of calculus, critical points are essential for understanding the behavior of functions. A critical point of a function occurs where the function's first derivative is equal to zero or the derivative does not exist. To find a critical point, you would calculate the derivative of the function and set it to zero, then solve for the variable in question.

Let's dissect the given function y = x^2 - x - 1. Following Step 1 of the exercise, taking its derivative yields y' = 2x - 1. At Step 2, we uncover the critical points by setting the derivative to zero and solving the resulting equation, leading us to x = 1/2. This point becomes a significant reference as we explore the function's overall behavior, and it will be crucial when we look for extreme values and absolute extrema.

Extreme Values of a Function

A function's extreme values represent the highest or lowest points within a certain interval or on a given domain. They're classified into two groups: local maxima and minima, which are the highest or lowest points compared to nearby values, and global (or absolute) maxima and minima, which are the highest or lowest points across the entire domain of the function.

Step 3 of our exercise applies the First Derivative Test, an effective tool in identifying these local extremes. By choosing numbers on either side of our critical point, x = 1/2, we can determine the function's behavior around that point. Since y'(x) changes from negative to positive as we pass x = 1/2, we conclude a local minimum exists at this critical point. The First Derivative Test gives us not only the locations of these turning points but also distinguishes between a local maxima and minima by studying the change in the sign of the derivative.

Absolute Extrema

The concept of absolute extrema, which includes both absolute maxima and minima, pertains to the highest and lowest values a function achieves on its entire domain. These values provide vital insights into the overall range and limits of the function, and they are particularly important in optimization problems.

In Step 4 of the solution, we investigate the absolute extremum by expanding our examination beyond local behavior to the entire set of possible values of x. Since the function y = x^2 - x - 1 is a polynomial, we know that it is continuous and differentiable everywhere, so we must consider both the critical points and the endpoints of the domain, which for polynomials extend to plus and minus infinity. By evaluating our function at the critical point x = 1/2, and considering the behavior of the function as x approaches infinity and negative infinity, we discover that y = -3/4 is indeed an absolute minimum – the lowest point on the entire domain of the function.