Absolute Minimum And Maximum Of F(x) = X⁴ - 50x² + 2 On [-4, 11]

Mr. Loba Loba · · 6 min read

Table Of Content

  1. Navigating the Landscape of f(x) = x⁴ - 50x² + 2
  2. Finding Critical Points: Where the Magic Happens
  3. Evaluating the Function at Critical Points and Endpoints
  4. Determining Absolute Minimum and Maximum Values
  5. Conclusion: The Extremes Unveiled

Hey guys! Let's dive into the fascinating world of functions and their extreme values. Today, we're going to explore the function f(x) = x⁴ - 50x² + 2 within the interval -4 ≤ x ≤ 11. Our mission? To find its absolute minimum and maximum values. Buckle up, because it's going to be a thrilling ride!

Before we jump into the calculations, let's get a feel for what this function looks like. The function f(x) = x⁴ - 50x² + 2 is a polynomial of degree 4, which means its graph will have a general W-shape or M-shape. The x⁴ term dominates for large values of x, causing the function to increase rapidly as x moves away from zero. The -50x² term introduces a dip in the middle, creating the possibility of local minima. The constant term +2 simply shifts the entire graph upwards by 2 units.

To find the absolute minimum and maximum values, we need to consider both the critical points of the function and the endpoints of the interval. Critical points are where the derivative of the function is either zero or undefined. In our case, since f(x) is a polynomial, its derivative will always be defined, so we only need to focus on where the derivative is zero. So guys, let's see how we can find it out!

Finding Critical Points: Where the Magic Happens

The critical points of a function are the key locations where the function's behavior might change direction – think of them as potential turning points where the function could reach a peak (maximum) or a valley (minimum). To pinpoint these critical points, we need to find where the derivative of our function, f'(x), equals zero. This is because at these points, the tangent line to the curve is horizontal, indicating a potential change in the function's slope from increasing to decreasing or vice versa.

Let's start by calculating the derivative of f(x) = x⁴ - 50x² + 2. Using the power rule, we get:

f'(x) = 4x³ - 100x

Now, we need to find the values of x for which f'(x) = 0. Setting the derivative to zero, we have:

4x³ - 100x = 0

We can factor out a 4x from the equation:

4x(x² - 25) = 0

This gives us three possible solutions:

  1. 4x = 0 => x = 0
  2. x² - 25 = 0 => x² = 25 => x = ±5

So, our critical points are x = -5, x = 0, and x = 5. Remember, we're only interested in the interval -4 ≤ x ≤ 11. While x = -5 falls outside this interval, the other two critical points, x = 0 and x = 5, are definitely within our domain of interest. These are the spots where our function might just be hitting its extreme highs or lows!

Evaluating the Function at Critical Points and Endpoints

Now that we've identified our critical points within the interval, the next crucial step is to evaluate the original function, f(x) = x⁴ - 50x² + 2, at these points. This will give us the function's value at these potential turning points. Additionally, we must also evaluate the function at the endpoints of our interval, which are x = -4 and x = 11. These endpoints are like the boundaries of our landscape, and the function might reach its highest or lowest point right at these edges.

Let's start by plugging in our critical points:

  1. x = 0:

    f(0) = (0)⁴ - 50(0)² + 2 = 2

  2. x = 5:

    f(5) = (5)⁴ - 50(5)² + 2 = 625 - 1250 + 2 = -623

Now, let's evaluate the function at the endpoints of our interval:

  1. x = -4:

    f(-4) = (-4)⁴ - 50(-4)² + 2 = 256 - 800 + 2 = -542

  2. x = 11:

    f(11) = (11)⁴ - 50(11)² + 2 = 14641 - 6050 + 2 = 8593

Okay, guys, we've got our function values at all the key locations! Now, let's organize these values and compare them to determine the absolute minimum and maximum.

Determining Absolute Minimum and Maximum Values

Alright, we've done the hard work of calculating the function's values at the critical points and endpoints. Now comes the moment of truth – identifying the absolute minimum and maximum values! Let's gather our results:

  • f(0) = 2
  • f(5) = -623
  • f(-4) = -542
  • f(11) = 8593

By simply comparing these values, we can easily spot the extremes. The smallest value is -623, which occurs at x = 5. This means that the absolute minimum value of the function f(x) on the interval -4 ≤ x ≤ 11 is -623.

On the other end of the spectrum, the largest value is 8593, which occurs at x = 11. Therefore, the absolute maximum value of the function f(x) on the given interval is 8593.

And there you have it! We've successfully navigated the function's landscape, identified its critical points, evaluated its values, and ultimately determined its absolute minimum and maximum values within the specified interval. Great job, guys!

Conclusion: The Extremes Unveiled

In this exhilarating exploration, we've successfully determined the absolute extremes of the function f(x) = x⁴ - 50x² + 2 within the interval -4 ≤ x ≤ 11. By meticulously calculating the derivative, identifying critical points, and evaluating the function at these points and the interval's endpoints, we've uncovered that the absolute minimum value is -623, occurring at x = 5, and the absolute maximum value is 8593, occurring at x = 11. This journey highlights the power of calculus in understanding the behavior of functions and pinpointing their extreme values – essential knowledge for various applications in mathematics, science, and engineering.

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