Hey there, math enthusiasts! Today, we're diving deep into the fascinating world of calculus to explore how the second derivative test helps us pinpoint local minimums and maximums of a function. We'll be focusing on the function f(x) = -2x³ + 30x² - 54x + 2, which, as the problem states, has one local minimum and one local maximum. Buckle up, because it's going to be an exciting ride!
Understanding the Second Derivative Test
Before we jump into the specifics of our function, let's quickly recap what the second derivative test is all about. Simply put, this test is a powerful tool that helps us determine whether a critical point of a function is a local minimum, a local maximum, or neither. Remember, critical points are those special x-values where the first derivative of the function is either zero or undefined. These points are the potential locations of our local extrema.
The second derivative of a function, as the name suggests, is the derivative of the first derivative. It tells us about the concavity of the function. If the second derivative is positive at a critical point, it means the function is concave up at that point, resembling a smile. This indicates a local minimum. Conversely, if the second derivative is negative, the function is concave down, resembling a frown, and we have a local maximum. If the second derivative is zero, the test is inconclusive, and we need to resort to other methods.
The beauty of the second derivative test lies in its efficiency. It often provides a straightforward way to classify critical points without having to analyze the function's behavior on either side of the point. This can save us a lot of time and effort, especially when dealing with more complex functions. So, with our theoretical foundation in place, let's get our hands dirty and apply the second derivative test to our function.
Step 1 Finding the First Derivative
The first step in our quest is to find the first derivative of f(x). This derivative, denoted as f'(x), will help us identify the critical points of the function, the potential locations of our local minimum and maximum. To find the derivative, we'll apply the power rule, which states that the derivative of xⁿ is nxⁿ⁻¹. Applying this rule to each term of our function, we get:
f'(x) = d/dx (-2x³ + 30x² - 54x + 2)
f'(x) = -6x² + 60x - 54
So, the first derivative of our function is a quadratic expression. This is a good sign because quadratic equations are relatively easy to solve, and we'll need to solve one in the next step to find our critical points. The first derivative, f'(x) = -6x² + 60x - 54, represents the slope of the tangent line to the graph of f(x) at any given point x. The points where this slope is zero are precisely the points where the function might have a local minimum or maximum. Therefore, setting the first derivative to zero is our next crucial step.
Step 2 Pinpointing Critical Points
Now that we have the first derivative, f'(x) = -6x² + 60x - 54, our next mission is to find the critical points. Remember, these are the x-values where the first derivative is either zero or undefined. In this case, our first derivative is a polynomial, which is defined for all real numbers. So, we only need to worry about where it's equal to zero.
To find the roots of the quadratic equation -6x² + 60x - 54 = 0, we can first simplify it by dividing both sides by -6, giving us:
x² - 10x + 9 = 0
Now, we can factor this quadratic equation. We're looking for two numbers that add up to -10 and multiply to 9. Those numbers are -1 and -9. So, we can factor the equation as:
(x - 1)(x - 9) = 0
Setting each factor to zero, we get our critical points:
x - 1 = 0 => x = 1
x - 9 = 0 => x = 9
So, we have two critical points x = 1 and x = 9. These are the potential locations of our local minimum and maximum. But how do we know which is which? That's where the second derivative test comes into play. We'll need to evaluate the second derivative at each of these points to determine the concavity of the function and thus identify the nature of the extrema.
Step 3 Unleashing the Second Derivative
With our critical points in hand, it's time to bring in the big guns the second derivative. The second derivative, f''(x), will tell us about the concavity of our function at these critical points. To find f''(x), we simply differentiate our first derivative, f'(x) = -6x² + 60x - 54, again using the power rule:
f''(x) = d/dx (-6x² + 60x - 54)
f''(x) = -12x + 60
Now we have a simple linear expression for the second derivative. This makes our job of evaluating it at the critical points much easier. The sign of f''(x) at a critical point will reveal whether we have a local minimum (positive), a local maximum (negative), or if the test is inconclusive (zero). So, let's evaluate f''(x) at our critical points, x = 1 and x = 9.
Step 4 Applying the Second Derivative Test
Now for the moment of truth! We'll plug our critical points, x = 1 and x = 9, into the second derivative, f''(x) = -12x + 60, to determine the concavity of the function at these points and thus identify whether they correspond to a local minimum or a local maximum.
Let's start with x = 1:
f''(1) = -12(1) + 60 = 48
Since f''(1) = 48 is positive, the function is concave up at x = 1. This means we have a local minimum at x = 1. Now, let's move on to x = 9:
f''(9) = -12(9) + 60 = -48
Since f''(9) = -48 is negative, the function is concave down at x = 9. This indicates that we have a local maximum at x = 9. We've successfully used the second derivative test to classify our critical points! But we're not quite done yet. We still need to find the actual values of the function at these local extrema.
Step 5 Finding the Values of the Extrema
We've identified the x-coordinates of our local minimum and maximum, but to complete the picture, we need to find the corresponding y-coordinates, which are the actual values of the function at these points. To do this, we simply plug our critical points, x = 1 and x = 9, back into the original function, f(x) = -2x³ + 30x² - 54x + 2.
Let's start with the local minimum at x = 1:
f(1) = -2(1)³ + 30(1)² - 54(1) + 2
f(1) = -2 + 30 - 54 + 2
f(1) = -24
So, the local minimum occurs at the point (1, -24), and the minimum value of the function is -24. Now, let's find the value of the local maximum at x = 9:
f(9) = -2(9)³ + 30(9)² - 54(9) + 2
f(9) = -1458 + 2430 - 486 + 2
f(9) = 488
Therefore, the local maximum occurs at the point (9, 488), and the maximum value of the function is 488. We've successfully found both the locations and the values of the local extrema of our function!
Conclusion: The Grand Finale
And there you have it, folks! We've successfully navigated the world of calculus and used the second derivative test to pinpoint the local minimum and maximum of the function f(x) = -2x³ + 30x² - 54x + 2. We found that the function has a local minimum at x = 1 with a value of -24, and a local maximum at x = 9 with a value of 488.
This journey demonstrates the power and elegance of calculus in helping us understand the behavior of functions. The second derivative test, in particular, is a valuable tool in our mathematical arsenal, allowing us to quickly and efficiently identify local extrema. Remember, guys, math isn't just about numbers and equations; it's about unlocking the secrets of the universe! So keep exploring, keep learning, and keep having fun with math!
Keywords: Second derivative test, local minimum, local maximum, critical points, concavity, first derivative, function analysis, extrema, calculus.