Absolute Max & Min Of F(x) = 5x^2 - 4x + 5 On [0, 10]

Oct 22, 2025 by ADMIN 54 views

Finding the Absolute Maximum and Minimum of f(x) = 5x² - 4x + 5 on the Interval [0, 10]

Hey guys! Let's dive into a classic calculus problem: finding the absolute maximum and minimum values of a function on a given interval. Today, we're going to tackle the function f(x) = 5x² - 4x + 5 on the interval 0 ≤ x ≤ 10. This is a fundamental concept in calculus, and mastering it will help you in various applications, from optimization problems to curve sketching. So, grab your calculators, and let's get started!

Understanding Absolute Extrema

Before we jump into the calculations, let's quickly recap what absolute maximum and minimum values are. The absolute maximum of a function on an interval is the largest value the function attains on that interval, while the absolute minimum is the smallest value. These values can occur at critical points (where the derivative is zero or undefined) or at the endpoints of the interval. Remember, guys, that finding these extrema is super useful in real-world applications where we need to optimize things like profit, cost, or efficiency. Think about it: businesses want to maximize profit and minimize costs, and this is exactly the kind of problem that calculus helps us solve. Understanding this foundational concept is crucial for anyone delving into calculus and its practical applications. To find the absolute extrema, we need to consider both the critical points within the interval and the endpoints of the interval. This ensures we capture the highest and lowest points the function reaches within the specified domain. This process involves finding the derivative of the function, setting it to zero to identify critical points, and then evaluating the function at these points and the interval's endpoints. By comparing these values, we can definitively determine the absolute maximum and minimum. It’s a systematic approach that guarantees we don’t miss any potential extrema.

Step 1: Find the Critical Points

The first step in finding the absolute extrema is to identify the critical points of the function within the given interval. Critical points are the points where the derivative of the function is either equal to zero or undefined. To find these points, we need to differentiate f(x) with respect to x. So, let's find the derivative, f'(x):

f(x) = 5x² - 4x + 5

Using the power rule, we get:

f'(x) = 10x - 4

Now, we need to set f'(x) equal to zero and solve for x:

10x - 4 = 0 10x = 4 x = 4/10 = 2/5 = 0.4

So, we have one critical point at x = 0.4. Since the derivative is a simple linear function, it's defined everywhere. Thus, x = 0.4 is our only critical point. But hold on, guys! We need to make sure this critical point lies within our interval, 0 ≤ x ≤ 10. Luckily, 0.4 is indeed within this interval, so we'll definitely need to consider it. This step is crucial because critical points indicate potential local maxima or minima, which can be candidates for the absolute extrema over the given interval. The derivative, being a measure of the function's rate of change, helps us pinpoint these points where the function's slope is either flat (derivative equals zero) or undefined. By identifying and evaluating these critical points, we narrow down the possible locations of the absolute maximum and minimum, making our search more efficient and accurate. It's like narrowing down the suspects in a detective story – we're getting closer to the solution!

Step 2: Evaluate the Function at Critical Points and Endpoints

Okay, now that we've found our critical point (x = 0.4), we need to evaluate the function f(x) at this point. We also need to evaluate the function at the endpoints of our interval, which are x = 0 and x = 10. This is because the absolute maximum and minimum values can occur either at a critical point within the interval or at one of the endpoints. So, let's calculate these values:

At x = 0 (Endpoint): f(0) = 5(0)² - 4(0) + 5 = 5
At x = 0.4 (Critical Point): f(0.4) = 5(0.4)² - 4(0.4) + 5 = 5(0.16) - 1.6 + 5 = 0.8 - 1.6 + 5 = 4.2
At x = 10 (Endpoint): f(10) = 5(10)² - 4(10) + 5 = 5(100) - 40 + 5 = 500 - 40 + 5 = 465

So, we have f(0) = 5, f(0.4) = 4.2, and f(10) = 465. This evaluation step is where we start to see the potential candidates for the absolute maximum and minimum emerge. By calculating the function's value at these key points, we create a set of values that we can directly compare to determine the extremes. Each value represents the function's height at a specific point along the interval, and these heights are what we're ultimately interested in. This is where the rubber meets the road – we're taking the theoretical framework and applying it to get concrete values that tell us about the function's behavior.

Step 3: Determine the Absolute Maximum and Minimum

Alright, we've done the heavy lifting! Now it's time to analyze the values we calculated in the previous step. We have:

f(0) = 5
f(0.4) = 4.2
f(10) = 465

By simply comparing these values, we can identify the absolute maximum and minimum. The absolute maximum is the largest value, and the absolute minimum is the smallest value.

So, it's clear that the largest value is 465, which occurs at x = 10. Therefore, the absolute maximum of f(x) on the interval [0, 10] is 465, and it occurs at x = 10. Similarly, the smallest value is 4.2, which occurs at x = 0.4. Therefore, the absolute minimum of f(x) on the interval [0, 10] is 4.2, and it occurs at x = 0.4. This final comparison step is crucial because it ties together all the previous steps into a conclusive answer. We're not just finding critical points or evaluating functions; we're using these tools to determine the ultimate highs and lows of the function within the specified domain. It's the culmination of our efforts, providing a clear and definitive understanding of the function's behavior.

Conclusion

And there you have it, guys! We've successfully found the absolute maximum and minimum values of the function f(x) = 5x² - 4x + 5 on the interval 0 ≤ x ≤ 10. The absolute maximum is 465, which occurs at x = 10, and the absolute minimum is 4.2, which occurs at x = 0.4. This process of finding absolute extrema is a fundamental skill in calculus, and it's used extensively in optimization problems. Remember, guys, the key steps are finding the critical points, evaluating the function at the critical points and endpoints, and then comparing the values to determine the maximum and minimum. Keep practicing, and you'll become a pro at these types of problems! Understanding how to find the absolute maximum and minimum of a function is not just an academic exercise; it's a powerful tool with real-world implications. Whether you're an engineer designing structures, an economist analyzing market trends, or a business owner optimizing operations, the ability to find extrema allows you to make informed decisions and achieve the best possible outcomes. So, keep honing your calculus skills, and you'll be well-equipped to tackle a wide range of optimization challenges. This process allows us to understand the bounds of a function's behavior within a specific range, providing crucial insights for decision-making and problem-solving in various fields. The systematic approach of identifying critical points, evaluating the function at these points and endpoints, and then comparing the values ensures that we capture the true extremes of the function, leading to accurate and reliable results.