Finding Min/Max Value Of F(x) = 3x^2 - 2x - 1 On [0, 2]

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Hey guys! Today, we're diving into a fundamental concept in calculus: finding the minimum and maximum values of a function within a given interval. This is super useful in many real-world applications, from optimizing business profits to designing efficient engineering systems. We'll break down the process step-by-step, using the example function f(x) = 3x^2 - 2x - 1 within the interval [0, 2]. Let's get started!

Understanding Extreme Values

Before we jump into the calculations, let's clarify what we mean by minimum and maximum values. These are collectively known as extreme values or extrema. A function's maximum value in a given interval is the highest point the function reaches, while the minimum value is the lowest point. These extreme values can occur at two types of points: critical points and endpoints.

  • Critical Points: These are points where the function's derivative (its slope) is either zero or undefined. At these points, the function can change direction, potentially forming a peak (maximum) or a valley (minimum).
  • Endpoints: These are the boundaries of the interval we're considering. The function's extreme values might occur at the very beginning or end of the interval.

Why are extreme values important? Identifying extreme values helps us understand the function's behavior. We can determine the function’s range within a certain interval and identify the input values that produce the highest or lowest outputs. This has a wide array of applications, such as determining the most cost-effective production level or finding the maximum height of a projectile. You'll use this in economics, physics, engineering, and many other fields.

Understanding these concepts is crucial for successfully finding the minimum and maximum values of a function. It's not just about the math; it's about understanding why we're doing it and what the results tell us. So, let's move on to the practical steps for finding these values.

Step 1: Finding the Derivative

The first step in finding the extreme values is to determine the derivative of the function. The derivative, often denoted as f'(x), represents the instantaneous rate of change of the function at any given point. In simpler terms, it tells us the slope of the line tangent to the function's graph at that point. For our example function, f(x) = 3x^2 - 2x - 1, we need to apply the power rule of differentiation. The power rule states that if we have a term of the form ax^n, its derivative is nax^(n-1).

Applying the power rule to each term in our function, we get:

  • The derivative of 3x^2 is 2 * 3x^(2-1) = 6x
  • The derivative of -2x is 1 * -2x^(1-1) = -2
  • The derivative of -1 (a constant) is 0

Combining these results, the derivative of f(x) is f'(x) = 6x - 2. Guys, this derivative is crucial, because it gives us the tool to find those critical points where the function's behavior might change. Think of it like a map that shows us the potential locations of peaks and valleys on our function's landscape. Without the derivative, we'd be wandering around blindly, hoping to stumble upon the extreme values. So, make sure you understand how to calculate derivatives – it's a key skill in calculus!

Why is the derivative important? The derivative represents the slope of the function at any given point. Critical points, where the derivative is zero or undefined, are potential locations of local maxima and minima. Understanding the derivative is paramount in finding these key points.

Step 2: Identifying Critical Points

Now that we have the derivative, f'(x) = 6x - 2, we can find the critical points. Remember, critical points occur where the derivative is either equal to zero or undefined. In our case, the derivative is a simple linear function, so it's defined for all values of x. This means we only need to find where the derivative equals zero.

So, we set f'(x) = 0 and solve for x:

6x - 2 = 0

Adding 2 to both sides:

6x = 2

Dividing both sides by 6:

x = 2/6 = 1/3

Therefore, we have one critical point at x = 1/3. What this tells us is that the function's slope is zero at this point, meaning it could be a local maximum, a local minimum, or a point of inflection (where the curve changes direction but doesn't form a peak or valley). To figure out which one it is, we'll need to evaluate the function at this point and compare it with other potential extreme values. Think of this critical point as a possible landmark on our function's graph – a place worth investigating further.

Why is identifying critical points crucial? Critical points are potential locations of local maxima and minima. They are points where the function's slope is either zero or undefined. Finding these points narrows down the locations where the function might have extreme values, making the overall search more efficient.

Step 3: Evaluating the Function at Critical Points and Endpoints

Now that we've found our critical point (x = 1/3), it's time to evaluate the function at this point and at the endpoints of our interval [0, 2]. This will give us the function's value at these key locations, allowing us to compare them and identify the minimum and maximum.

Let's start by evaluating the function at the critical point, x = 1/3:

f(1/3) = 3(1/3)^2 - 2(1/3) - 1 = 3(1/9) - 2/3 - 1 = 1/3 - 2/3 - 1 = -4/3

So, the function's value at the critical point is -4/3. Now, we need to evaluate the function at the endpoints of the interval:

  • At x = 0:

    f(0) = 3(0)^2 - 2(0) - 1 = -1

  • At x = 2:

    f(2) = 3(2)^2 - 2(2) - 1 = 3(4) - 4 - 1 = 12 - 4 - 1 = 7

Okay, so we have three values to compare: f(1/3) = -4/3, f(0) = -1, and f(2) = 7. These values are the potential candidates for the minimum and maximum values of the function within the given interval. Evaluating the function at critical points and endpoints is a crucial step because it provides the specific values we need to make a direct comparison. Without this step, we'd be left with just potential locations, but no actual values to work with.

Step 4: Determining the Minimum and Maximum Values

Alright, we've done the hard work! We've calculated the function's values at the critical point (x = 1/3) and the endpoints of the interval (x = 0 and x = 2). Now comes the final step: comparing these values to determine the minimum and maximum. We have:

  • f(1/3) = -4/3 ≈ -1.33
  • f(0) = -1
  • f(2) = 7

By simply looking at these values, it's clear that the smallest value is -4/3, which occurs at x = 1/3. Therefore, the minimum value of the function f(x) = 3x^2 - 2x - 1 in the interval [0, 2] is -4/3.

Similarly, the largest value is 7, which occurs at x = 2. So, the maximum value of the function in the interval [0, 2] is 7. And that's it! We've successfully found both the minimum and maximum values. The minimum value here occurs at the critical point while the maximum value occurs at the endpoint. This highlights the importance of checking both types of points. Remember, the process of comparing values is the culmination of all our previous work. It's the moment where we actually answer the question we set out to solve.

Why is it so important to compare the values? Comparing the function values at critical points and endpoints allows us to definitively identify the absolute minimum and maximum values within the specified interval. This comparison is the ultimate step in determining the extreme values of the function.

Visualizing the Solution (Optional)

To solidify your understanding, it's helpful to visualize the solution. If you were to graph the function f(x) = 3x^2 - 2x - 1, you'd see a parabola opening upwards. The interval [0, 2] would be a section of this parabola. The minimum point would be the lowest point on the parabola within that interval, which we found to be at x = 1/3. The maximum point would be the highest point within the interval, which is at the endpoint x = 2.

Visualization helps in two main ways:

  1. Confirmation: It provides a visual confirmation of our calculations. We can see that our minimum and maximum values make sense in the context of the graph.
  2. Intuition: It strengthens our intuition about how functions behave. We can see how critical points and endpoints play a role in determining extreme values.

Conclusion

So guys, finding the minimum and maximum values of a function is a powerful technique with many practical applications. By following these steps – finding the derivative, identifying critical points, evaluating the function at critical points and endpoints, and comparing the values – you can confidently determine the extreme values of a function within a given interval. Remember to understand the why behind the steps, not just the how. This understanding will help you apply these concepts in various contexts and solve real-world problems. Keep practicing, and you'll become a master of finding extreme values!