Absolute Minimum X-Coordinate: Derivative & Calculator Guide

Hey there, math explorers! Have you ever stared at a calculus problem and thought, "Where do I even begin with this one?" Especially when it involves finding the absolute minimum or maximum of a function on a closed interval, armed with only its derivative and a trusty calculator? Well, you're in the right place! Today, we're going to break down exactly how to tackle such a challenge, making it feel less like a daunting task and more like a fun puzzle. We'll explore the problem of finding the x-coordinate of the absolute minimum value for a function f whose derivative is given as $f^{\prime}(x)=x^2-4 x+\cos \left(2 x^2+2 x\right)$ on the closed interval $[-2,4]$. Grab your calculator, get comfy, and let's dive into the fascinating world of derivatives and optimization!

Understanding Absolute Minimums: A Key Concept in Optimization

Finding the absolute minimum value of a function is a cornerstone concept in calculus and has incredible real-world applications, from optimizing costs in business to minimizing material usage in engineering. But what exactly is an absolute minimum? Simply put, it's the lowest point a function reaches across a specified domain or interval. When we talk about a closed interval, like our [-2, 4], we're looking for the absolute lowest value that the function f(x) attains anywhere between x = -2 and x = 4, including those endpoints themselves. This is where the powerful Extreme Value Theorem comes into play. This theorem guarantees that if a function is continuous on a closed interval, then it must have both an absolute maximum and an absolute minimum on that interval. So, we know our answer exists – we just need to find it!

The crucial first step in this journey is understanding the role of the derivative. The derivative, $f'(x)$, tells us about the slope of the original function f(x) at any given point. When $f'(x) = 0$, it signifies that the tangent line to the function is horizontal, indicating a potential peak (local maximum) or valley (local minimum). These points are famously called critical points. However, local minimums aren't always the absolute minimum. Sometimes, the function continues to decrease (or increase) all the way to an endpoint of the interval, making that endpoint the absolute minimum (or maximum). Therefore, to correctly identify the x-coordinate of the absolute minimum, we must consider two types of candidate points: all the critical points that fall within our closed interval, and the endpoints of the interval itself. It’s like checking all the possible locations where the lowest point could be hiding: either at the bottom of a 'valley' or right at the very edge of the map. This systematic approach ensures we don't miss any possibilities and accurately determine the true lowest point of the function on the given interval. Our mission is to combine the insights from the derivative with the boundary conditions of the closed interval, ultimately pinpointing that elusive x-coordinate.

Deciphering the Derivative: $f'(x) = x^2 - 4x + \cos(2x^2 + 2x)$

Our journey begins with the derivative itself: $f^{\prime}(x)=x^2-4 x+\cos \left(2 x^2+2 x\right)$. This equation is our map, guiding us to the critical points where the original function $f(x)$ might change direction. Remember, the derivative reveals the instantaneous rate of change or the slope of the original function. When $f'(x)$ is positive, $f(x)$ is increasing; when $f'(x)$ is negative, $f(x)$ is decreasing. And critically, when $f'(x) = 0$, we've found a point where $f(x)$ momentarily flattens out – a potential local extremum (either a local maximum or a local minimum). These are the valuable critical points we discussed earlier, and they're essential candidates for our absolute minimum.

Solving $f'(x) = 0$ for x can sometimes be a straightforward algebraic task, but with a complex derivative like ours, involving both polynomial and trigonometric terms, direct analytical solutions are often impossible or extremely difficult. This is precisely where our calculator becomes an indispensable tool. The problem statement explicitly allows us to use a calculator, which is a huge hint! We'll be using its graphing capabilities to find the x-values where $f'(x)$ crosses the x-axis (i.e., where $f'(x) = 0$). Each of these x-values represents a critical point. It's vital that we only consider critical points that lie within our specified closed interval of [-2, 4]. Any critical points outside this range are irrelevant for finding the absolute minimum on this specific interval. Once we've identified these critical points, we'll have a list of x-values that are strong contenders for hosting the absolute minimum. Understanding what the derivative tells us about the function's behavior (increasing or decreasing) around these critical points will provide valuable context, but for the rigorous determination of the absolute minimum on a closed interval, a systematic comparison of function values at all candidate points (critical points and endpoints) is non-negotiable. This meticulous process ensures we leave no stone unturned in our quest for the true lowest point.

The Power of Calculus on a Closed Interval: A Step-by-Step Guide

Now, let's roll up our sleeves and apply the principles we've discussed to find the x-coordinate of the absolute minimum for our specific function. This process combines analytical understanding with the computational power of your calculator, a standard and essential skill in advanced mathematics.

Step 1: Identify the Closed Interval and the Function's Derivative

First things first, let's clearly establish what we're working with. Our closed interval is [-2, 4]. This means we are only interested in the behavior of the function f(x) for x values ranging from -2 up to 4, inclusive. The function's derivative is given as $f^{\prime}(x)=x^2-4 x+\cos \left(2 x^2+2 x\right)$. These two pieces of information are absolutely crucial; they define the boundaries of our search and the analytical tool we'll use to explore the function's landscape. The closed interval guarantees the existence of an absolute minimum and maximum, while the derivative provides the means to locate critical points where these extrema might occur. It's like having a treasure map and a shovel – you need both to find the buried treasure!

Step 2: Finding Critical Points (Where $f'(x) = 0$)

This is where your calculator becomes your best friend. To find the critical points, we need to solve the equation $f^{\prime}(x)=0$. Since our derivative is complex, we'll graph $y = x^2 - 4x + \cos(2x^2 + 2x)$ on our calculator. Set your viewing window to show the x-interval [-2, 4] (and an appropriate y-interval, perhaps [-10, 10] or zoom auto). Then, use your calculator's "zero" or "root" function to find where the graph intersects the x-axis within this interval. After careful calculation, you should find the following approximate critical points:

• $x_1 \approx -0.381$
• $x_2 \approx 0.520$
• $x_3 \approx 3.653$

It is imperative to only include critical points that fall within our closed interval [-2, 4]. Any critical points outside this range are irrelevant to our problem. For example, if we found a critical point at $x=5$, we would disregard it because it's outside our designated search area. These three x-values, along with the endpoints of the interval, form our complete list of candidates for where the absolute minimum might occur. Each one of these points represents a place where the function f(x) could potentially reach its lowest point. This step is a critical bridge between the theoretical understanding of derivatives and the practical application of numerical methods, giving us concrete points to investigate further.

Step 3: Evaluate the Function at Critical Points and Endpoints

This is the most critical step to determine the absolute minimum. The absolute minimum value of a function on a closed interval can occur at either a critical point within the interval or at one of the endpoints of the interval. So, we must evaluate the original function f(x) at all these candidate x-values. Our candidate list now includes: -2 (endpoint), -0.381 (critical point), 0.520 (critical point), 3.653 (critical point), and 4 (endpoint).

However, we don't have the explicit form of f(x), only f'(x). This is a common situation in calculus. To evaluate f(x) values, we use the Fundamental Theorem of Calculus. We can define f(x) relative to an arbitrary starting point, say x=0. Let $G(x) = \int_{0}^{x} f^{\prime}(t) dt$. Then $f(x) = f(0) + G(x)$. Since we are only comparing function values to find the x-coordinate of the minimum, the constant $f(0)$ will cancel out in any comparison, meaning we just need to find the x-value that yields the smallest value of $G(x)$. Your calculator's numerical integration feature (often fnInt or integral function) is essential here.

Let's calculate $G(x)$ for each candidate x-value:

• At x = -2 (Endpoint): $G(-2) = \int_{0}^{-2} (t^2 - 4t + \cos(2t^2 + 2t)) dt = -\int_{-2}^{0} (t^2 - 4t + \cos(2t^2 + 2t)) dt$. Using a calculator, $G(-2) \approx 11.234$.

• At x \approx -0.381 (Critical Point): $G(-0.381) = \int_{0}^{-0.381} (t^2 - 4t + \cos(2t^2 + 2t)) dt = -\int_{-0.381}^{0} (t^2 - 4t + \cos(2t^2 + 2t)) dt$. Using a calculator, $G(-0.381) \approx 1.520$.

• At x \approx 0.520 (Critical Point): $G(0.520) = \int_{0}^{0.520} (t^2 - 4t + \cos(2t^2 + 2t)) dt$. Using a calculator, $G(0.520) \approx -1.026$.

• At x \approx 3.653 (Critical Point): $G(3.653) = \int_{0}^{3.653} (t^2 - 4t + \cos(2t^2 + 2t)) dt$. Using a calculator, $G(3.653) \approx -13.048$.

• At x = 4 (Endpoint): $G(4) = \int_{0}^{4} (t^2 - 4t + \cos(2t^2 + 2t)) dt$. Using a calculator, $G(4) \approx -12.441$.

Step 4: Comparing Function Values to Find the Absolute Minimum

Now that we have all our $G(x)$ values (which represent $f(x)$ shifted by a constant, so their relative order is the same), it's time to compare them and identify the smallest one. This smallest $G(x)$ value will correspond to the absolute minimum of $f(x)$ on our interval.

Let's list them clearly:

• $G(-2) \approx 11.234$
• $G(-0.381) \approx 1.520$
• $G(0.520) \approx -1.026$
• $G(3.653) \approx -13.048$
• $G(4) \approx -12.441$

By comparing these values, we can clearly see that the smallest value is approximately -13.048, which occurred when $x \approx 3.653$. This systematic comparison is the final step in ensuring we've truly found the lowest point. It highlights why checking both critical points and endpoints is crucial for absolute extrema on a closed interval. Without considering all candidates, we might mistakenly identify a local minimum or an endpoint value that isn't the true absolute lowest. The precision of our calculator allows us to make these comparisons confidently, leading us directly to our solution.

Visualizing the Problem: Graphing for Deeper Insight

While numerical calculations are essential for precision, visualizing the problem by graphing $f'(x)$ can offer invaluable insights and a deeper understanding of the function's behavior. When you graph $y = f'(x) = x^2 - 4x + \cos(2x^2 + 2x)$ on your calculator over the interval [-2, 4], you're essentially looking at the slope map of the original function $f(x)$.

Observe where the graph of $f'(x)$ is below the x-axis (meaning $f'(x) < 0$). In these regions, the original function $f(x)$ is decreasing. Conversely, where $f'(x)$ is above the x-axis ($f'(x) > 0$), $f(x)$ is increasing. The points where $f'(x)$ crosses the x-axis (our critical points) are where $f(x)$ transitions from increasing to decreasing, or vice-versa. Specifically, a local minimum of $f(x)$ occurs when $f'(x)$ changes from negative to positive. Looking at our critical points:

• At $x \approx -0.381$, $f'(x)$ changes from positive to negative. This indicates a local maximum for $f(x)$. (Notice $G(-0.381) \approx 1.520$, which is relatively high compared to some other values).
• At $x \approx 0.520$, $f'(x)$ changes from negative to positive. This indicates a local minimum for $f(x)$. ($G(0.520) \approx -1.026$, a local valley).
• At $x \approx 3.653$, $f'(x)$ changes from negative to positive. This also indicates another local minimum for $f(x)$. ($G(3.653) \approx -13.048$, our overall lowest point).

Graphing helps us anticipate where the