Absolute Max/Min Of F(x) = 3 - 7x^2 On [-5, 1]

Hey guys! Today, we're diving into a classic calculus problem: finding the absolute maximum and minimum values of a function on a given interval. Specifically, we'll be working with the function f(x) = 3 - 7x² on the interval -5 ≤ x ≤ 1. This might seem a bit daunting at first, but don't worry, we'll break it down step by step so it's super easy to understand. So, grab your thinking caps, and let's get started!

Understanding Absolute Extrema

Before we jump into the calculations, let's quickly recap what we mean by absolute maximum and absolute minimum. Think of it this way: the absolute maximum is the highest point the function reaches on the entire interval, and the absolute minimum is the lowest point. These points are also known as the absolute extrema of the function.

Now, where can these absolute extrema occur? There are two main possibilities:

1. Critical Points: These are the points where the derivative of the function is either zero or undefined. They represent potential turning points of the function, where it might switch from increasing to decreasing or vice versa. Key concept here is that at critical points the slope of the tangent line to the function's graph is either zero (horizontal tangent) or undefined (vertical tangent or a cusp).
2. Endpoints: The absolute maximum or minimum could also occur at the endpoints of the interval. This is because the function might be increasing or decreasing right up to the edge of the interval. Consider a function that's steadily decreasing across the interval; clearly, its absolute maximum will be at the left endpoint, and its absolute minimum will be at the right endpoint. Always remember to check the endpoints, as they can sometimes be overlooked, leading to incorrect results.

So, our strategy is clear: we'll find the critical points of the function within the given interval, evaluate the function at these critical points and at the endpoints, and then compare the values to determine the absolute maximum and minimum.

Finding the Critical Points

Alright, let's get our hands dirty with some calculus! The first step in finding the critical points is to find the derivative of our function, f(x) = 3 - 7x². Using the power rule, we get:

f'(x) = -14x

Now, to find the critical points, we need to find where this derivative is equal to zero or undefined. In this case, the derivative is a simple linear function, and it's defined for all values of x. So, we just need to solve for when f'(x) = 0:

-14x = 0

x = 0

Great! We found one critical point: x = 0. Since 0 is within our interval of -5 ≤ x ≤ 1, we'll need to consider it.

Why are critical points so crucial? They're the potential locations where our function transitions from increasing to decreasing (or vice versa). Imagine a roller coaster; the peaks and valleys are where the ride changes direction, much like critical points mark changes in a function's behavior. This is why we meticulously identify and analyze these points when seeking extrema. Moreover, understanding the nature of the critical point—whether it’s a local maximum, a local minimum, or neither—can provide a deeper insight into the function’s overall behavior. Techniques like the first derivative test or the second derivative test can help classify these critical points, giving us a more detailed picture of the function's graph.

Evaluating the Function

Now that we have our critical point (x = 0) and our endpoints (x = -5 and x = 1), we need to evaluate the original function, f(x) = 3 - 7x², at these points. This will tell us the y-values corresponding to these x-values.

Let's start with the critical point:

f(0) = 3 - 7(0)² = 3

So, at x = 0, the function value is 3.

Next, let's evaluate the function at the endpoints:

f(-5) = 3 - 7(-5)² = 3 - 7(25) = 3 - 175 = -172

So, at x = -5, the function value is -172.

And finally, at x = 1:

f(1) = 3 - 7(1)² = 3 - 7 = -4

So, at x = 1, the function value is -4.

By evaluating the function at these key points, we're essentially taking snapshots of the function's height across the interval. These snapshots are crucial because they help us identify the highest and lowest points, which directly correspond to the absolute maximum and minimum values. Think of it as climbing a mountain; to find the highest and lowest points, you need to check the peaks, the valleys, and the very edges of your climbing area. This is precisely what we're doing mathematically by evaluating the function at critical points and endpoints.

Determining Absolute Maximum and Minimum

Okay, we've done the hard work! We have the function values at the critical point and endpoints:

• f(0) = 3
• f(-5) = -172
• f(1) = -4

Now, it's just a matter of comparing these values to find the absolute maximum and minimum. Remember, the absolute maximum is the largest value, and the absolute minimum is the smallest value.

Looking at our values, we can see that the largest value is 3, which occurs at x = 0. So, the absolute maximum value is 3, and it occurs at x = 0.

And the smallest value is -172, which occurs at x = -5. So, the absolute minimum value is -172, and it occurs at x = -5.

Understanding why this method works is just as important as the calculation itself. We're leveraging the fact that continuous functions on closed intervals are guaranteed to have both an absolute maximum and an absolute minimum. By checking critical points and endpoints, we're systematically covering all potential locations where these extrema could exist. The critical points are key because they represent points where the function's rate of change is zero, indicating a potential peak or valley. The endpoints, on the other hand, are crucial because the function might be increasing or decreasing right up to the boundary of the interval, making an endpoint the location of an extremum.

Conclusion

And there you have it! We've successfully found the absolute maximum and minimum values of the function f(x) = 3 - 7x² on the interval -5 ≤ x ≤ 1. To summarize:

• The absolute maximum value is 3, and it occurs at x = 0.
• The absolute minimum value is -172, and it occurs at x = -5.

This problem is a classic example of how calculus can be used to find the extreme values of a function. The key steps are:

1. Find the derivative of the function.
2. Find the critical points by setting the derivative equal to zero and solving for x.
3. Evaluate the function at the critical points and the endpoints of the interval.
4. Compare the values to determine the absolute maximum and minimum.

By following these steps, you can tackle a wide range of optimization problems in calculus. Remember, understanding the underlying concepts—like what critical points represent and why we check endpoints—is just as crucial as the calculations themselves. So, keep practicing, keep exploring, and you'll become a calculus whiz in no time!

I hope this explanation was helpful and clear! If you have any questions or want to try out more examples, feel free to ask. Happy calculating, everyone! Remember that math, like any skill, improves with practice. Don't be afraid to tackle challenging problems, and always seek to understand the 'why' behind the 'how'. This approach will not only enhance your problem-solving abilities but also deepen your appreciation for the elegance and power of calculus.