Absolute Min And Max Consider F(x) = X^4 - 72x^2 + 10
Hey guys! Let's dive into a fascinating problem in calculus where we need to find the absolute minimum and maximum values of a function. We'll break it down step by step, making sure everyone understands the process. Today, we're tackling the function f(x) = x^4 - 72x^2 + 10, defined over the interval -5 ≤ x ≤ 13. Our mission is to pinpoint the absolute minimum and absolute maximum values this function attains within this interval. Buckle up, because it's going to be an exciting ride!
Understanding the Problem
Before we jump into the calculations, let's make sure we understand what the question is asking. We're given a function, f(x) = x^4 - 72x^2 + 10, which is a polynomial function. Polynomials are nice and smooth, meaning they don't have any abrupt jumps or breaks. We're also given an interval, -5 ≤ x ≤ 13, which means we're only interested in the behavior of the function within this range of x-values.
The absolute minimum value is the lowest point the function reaches within the given interval, and the absolute maximum value is the highest point. These points can occur either at the critical points of the function (where the derivative is zero or undefined) or at the endpoints of the interval. So, our strategy will be to find these critical points, evaluate the function at these points and the endpoints, and then compare the values to determine the absolute minimum and maximum.
Why This Matters
Finding absolute minima and maxima isn't just an abstract mathematical exercise. It has tons of real-world applications. Imagine you're an engineer designing a bridge. You need to know the maximum stress the bridge can withstand. Or, suppose you're an economist trying to predict the highest possible profit for a company. These kinds of problems often boil down to finding the extreme values of a function. So, mastering this skill is super practical!
Step 1: Finding the Critical Points
Okay, let's get our hands dirty with some calculus! The first step in finding the absolute minimum and maximum values is to find the critical points of the function. Remember, critical points are the points where the derivative of the function is either equal to zero or undefined. Since our function is a polynomial, it's defined everywhere, so we only need to worry about where the derivative is zero.
To find the critical points, we first need to find the derivative of f(x). Using the power rule, we have:
f'(x) = 4x^3 - 144x
Now, we need to set f'(x) equal to zero and solve for x:
4x^3 - 144x = 0
We can factor out a 4x from the equation:
4x(x^2 - 36) = 0
This gives us three possible solutions:
- 4x = 0 which means x = 0
- x^2 - 36 = 0 which means x^2 = 36, so x = ±6
So, our critical points are x = -6, x = 0, and x = 6. However, we need to consider only those critical points that lie within our interval -5 ≤ x ≤ 13. Notice that x = -6 is outside our interval, so we can discard it. This leaves us with two critical points within the interval: x = 0 and x = 6. Awesome! We're making progress.
A Deeper Dive into Derivatives
Why are we even looking at the derivative? Well, the derivative tells us about the slope of the function at any given point. At the critical points, the slope is either zero (a horizontal tangent line) or undefined (a vertical tangent line or a sharp corner). These are the places where the function can potentially change direction, going from increasing to decreasing or vice versa. That's why critical points are crucial for finding the extrema (minima and maxima) of a function.
Step 2: Evaluating the Function
Now that we've found our critical points within the interval (x = 0 and x = 6), we need to evaluate the original function, f(x) = x^4 - 72x^2 + 10, at these points. We also need to evaluate the function at the endpoints of our interval, which are x = -5 and x = 13. This will give us a set of values that we can compare to find the absolute minimum and maximum.
Let's start with the endpoints:
- f(-5) = (-5)^4 - 72(-5)^2 + 10 = 625 - 72(25) + 10 = 625 - 1800 + 10 = -1165
- f(13) = (13)^4 - 72(13)^2 + 10 = 28561 - 72(169) + 10 = 28561 - 12168 + 10 = 16303
Now, let's evaluate the function at the critical points:
- f(0) = (0)^4 - 72(0)^2 + 10 = 0 - 0 + 10 = 10
- f(6) = (6)^4 - 72(6)^2 + 10 = 1296 - 72(36) + 10 = 1296 - 2592 + 10 = -1286
So, we have the following values:
- f(-5) = -1165
- f(13) = 16303
- f(0) = 10
- f(6) = -1286
The Importance of Endpoints
You might be wondering why we need to check the endpoints of the interval. Well, the absolute minimum or maximum value of a function can occur at a critical point, but it can also occur at an endpoint. Think about a line segment. The highest or lowest point will always be at one of the endpoints, unless the line is horizontal. Similarly, for more complex functions, the endpoints are potential locations for extreme values.
Step 3: Determining the Absolute Minimum and Maximum
Alright, we've done the hard work! We've found the critical points, evaluated the function at these points and the endpoints, and now we have a set of values to compare. Let's recap our values:
- f(-5) = -1165
- f(13) = 16303
- f(0) = 10
- f(6) = -1286
To find the absolute minimum value, we simply look for the smallest value among these. That's f(6) = -1286. So, the absolute minimum value of the function on the interval -5 ≤ x ≤ 13 is -1286.
To find the absolute maximum value, we look for the largest value. That's f(13) = 16303. Therefore, the absolute maximum value of the function on the given interval is 16303.
Putting It All Together
We've successfully navigated through this problem! We started by understanding the question and the concepts of absolute minimum and maximum values. We then found the critical points of the function by taking the derivative and setting it equal to zero. We evaluated the function at the critical points and the endpoints of the interval. Finally, we compared the values to determine the absolute minimum and maximum.
Conclusion
So, to answer the original question, the function f(x) = x^4 - 72x^2 + 10 on the interval -5 ≤ x ≤ 13 has an absolute minimum value of -1286 and an absolute maximum value of 16303.
This problem demonstrates a powerful technique in calculus for finding the extreme values of a function. Remember the steps: find the critical points, evaluate the function at the critical points and endpoints, and then compare the values. This approach can be applied to a wide range of problems, from optimization in engineering to maximizing profits in business. Keep practicing, and you'll become a pro at finding minima and maxima! You got this!
