Finding The Maximum Value Of F(x) = E^x - 2 On [0, 2] Using Calculus

Hey guys! Today, let's dive into a classic calculus problem: finding the maximum value of a function on a given interval. We'll tackle the function f(x) = e^x - 2 on the interval [0, 2]. This is a super common type of problem you'll see in calculus, so understanding how to solve it is really important. We'll break down the steps and make sure you get the hang of it. So, grab your thinking caps, and let's get started!

Understanding the Problem

Before we jump into the calculations, let's make sure we understand what the problem is asking. We have a function, f(x) = e^x - 2, which is an exponential function with a vertical shift. We want to find the largest value this function reaches when x is between 0 and 2, inclusive. This means we're looking for the highest point on the graph of the function within that specific interval.

The interval [0, 2] is crucial because it tells us the boundaries we need to consider. We're not interested in what the function does outside of this range. Think of it like zooming in on a specific section of the graph. Now, how do we actually find this maximum value? That's where calculus comes in! Calculus gives us the tools to analyze how functions change and find their highest and lowest points. We'll be using derivatives, which tell us about the slope of the function, and critical points, which are potential locations of maxima and minima. Don't worry if these terms sound intimidating – we'll walk through them step by step. By the end of this article, you'll be a pro at finding maximum values on intervals! So, keep reading, and let's unlock the secrets of this problem together.

Calculus Approach: Finding Critical Points

The calculus approach to finding the maximum value involves a few key steps, primarily focusing on finding critical points. Critical points are those points where the derivative of the function is either zero or undefined. These points are crucial because they are potential locations for local maxima or minima. In simpler terms, they are the spots where the function might change direction – going from increasing to decreasing (a maximum) or vice versa (a minimum). So, let's dive into the steps we need to take. First up, we need to find the derivative of our function, f(x) = e^x - 2. Remember, the derivative tells us the instantaneous rate of change of the function at any given point. The derivative of e^x is simply e^x, and the derivative of a constant (-2 in this case) is zero. So, the derivative of f(x), denoted as f'(x), is just e^x. Easy peasy, right? Now that we have the derivative, the next step is to find the critical points. We do this by setting the derivative equal to zero and solving for x. So, we have the equation e^x = 0. Here's a little trick to keep in mind: the exponential function e^x is always positive. It never touches or crosses the x-axis, meaning it never equals zero. This tells us that there are no solutions to the equation e^x = 0. Therefore, our function f(x) = e^x - 2 has no critical points where the derivative is zero. But wait, we're not done yet! Remember, critical points can also occur where the derivative is undefined. However, the exponential function e^x is defined for all real numbers. So, there are no critical points where the derivative is undefined either. This might seem a bit puzzling – if we don't have any critical points, how do we find the maximum value? Well, this is where the interval [0, 2] comes back into play. Since we don't have any critical points within the interval, the maximum value must occur at one of the endpoints of the interval. This is a super important concept in calculus: when finding the maximum or minimum value of a function on a closed interval, you always need to check the endpoints, even if you find critical points! So, in the next section, we'll evaluate the function at the endpoints of our interval and see which one gives us the maximum value. Keep those thinking caps on, guys – we're almost there!

Evaluating Endpoints

Since we found that our function f(x) = e^x - 2 has no critical points within the interval [0, 2], the maximum value must occur at one of the endpoints. This is a key concept in optimization problems in calculus. When dealing with a continuous function on a closed interval, the extreme values (maximum and minimum) will either occur at critical points within the interval or at the endpoints of the interval. In our case, we've already established that there are no critical points within the interval [0, 2]. So, all that's left to do is to evaluate the function at the endpoints and compare the results. The endpoints of our interval are x = 0 and x = 2. Let's start by evaluating the function at x = 0. We have f(0) = e^0 - 2. Remember that any number raised to the power of 0 is 1, so e^0 = 1. Therefore, f(0) = 1 - 2 = -1. Now, let's evaluate the function at x = 2. We have f(2) = e^2 - 2. This is where things get a little more interesting. e^2 is the square of the base of the natural logarithm, which is approximately 2.718. So, e^2 is approximately 2.718 squared, which is roughly 7.389. Therefore, f(2) = e^2 - 2 is approximately 7.389 - 2 = 5.389. Now we can compare the values we found at the endpoints: f(0) = -1 and f(2) ≈ 5.389. It's clear that f(2) is significantly larger than f(0). This tells us that the maximum value of the function f(x) = e^x - 2 on the interval [0, 2] occurs at x = 2, and the maximum value is e^2 - 2. Guys, isn't this cool? We've used calculus to pinpoint the highest point of the function within our specified range. In the next section, we'll wrap up our findings and select the correct answer from the options provided. So, stick with me, and let's nail this problem!

Determining the Maximum Value

Alright, we've done the heavy lifting! We've analyzed the function f(x) = e^x - 2, found its derivative, identified that it has no critical points within the interval [0, 2], and evaluated the function at the endpoints of the interval. Now, it's time to put it all together and determine the maximum value. Remember, we calculated f(0) = -1 and f(2) = e^2 - 2. Since e^2 - 2 is approximately 5.389, which is much greater than -1, we can confidently say that the maximum value of the function on the interval [0, 2] is e^2 - 2. Now, let's revisit the options given in the problem: A) 0, B) e - 2, C) e^2 - 2, D) 2. We can see that option C, e^2 - 2, matches our calculated maximum value perfectly! This is awesome – we've successfully navigated through the problem and found the correct answer using calculus principles. Guys, give yourselves a pat on the back! We've tackled a common type of calculus problem, and you've seen how the steps come together to lead us to the solution. From finding the derivative to evaluating the function at endpoints, each step plays a crucial role in determining the maximum value. Now, before we wrap up completely, let's just take a moment to reflect on what we've learned. This type of problem highlights the power of calculus in optimization – finding the best possible outcome (in this case, the maximum value) within certain constraints (the interval [0, 2]). These optimization techniques are used in a wide range of fields, from engineering and economics to computer science and physics. So, the skills you're developing here are not just for passing your calculus class; they're valuable tools for problem-solving in the real world. In the next (and final!) section, we'll summarize our solution and recap the key takeaways from this problem. So, let's keep the momentum going and finish strong!

Conclusion and Answer

Okay, guys, let's bring it all home! We started with the problem of finding the maximum value of the function f(x) = e^x - 2 on the interval [0, 2]. We've journeyed through the calculus landscape, using derivatives and endpoint evaluation to arrive at our solution. Let's quickly recap the key steps we took. First, we found the derivative of the function, f'(x) = e^x. Then, we looked for critical points by setting the derivative equal to zero and found that there were none within our interval. This meant that the maximum value had to occur at one of the endpoints. We evaluated the function at x = 0 and x = 2, obtaining f(0) = -1 and f(2) = e^2 - 2. Comparing these values, we determined that the maximum value of the function on the interval [0, 2] is e^2 - 2. Now, let's circle back to the original question and the answer choices. The question asked: "Qual é o valor máximo da função f(x) = e^x - 2 no intervalo de 0 a 2, e como podemos determinar esse valor utilizando cálculo?" (What is the maximum value of the function f(x) = e^x - 2 on the interval from 0 to 2, and how can we determine this value using calculus?) And the answer choices were: A) 0, B) e - 2, C) e^2 - 2, D) 2. Based on our calculations and analysis, the correct answer is C) e^2 - 2. We not only found the maximum value, but we also demonstrated the calculus techniques used to determine it. This problem highlights the importance of understanding derivatives, critical points, and endpoint evaluation when finding extreme values of functions on closed intervals. These are fundamental concepts in calculus and have wide-ranging applications in various fields. So, guys, I hope this walkthrough has been helpful and has solidified your understanding of this type of problem. Remember to practice these techniques, and you'll be well-equipped to tackle similar challenges in the future. Keep up the great work, and happy calculating!