Finding Stationary Points And Values Of F(x) = X³ - 6x² + 5

Let's dive into finding the stationary points and values of the function f(x) = x³ - 6x² + 5. This is a classic calculus problem, and we'll break it down step by step so it's super clear. We will explore how to find stationary points, calculate stationary values, and classify the nature of these points.

a. Determining Stationary Points of the Function f(x)

To determine stationary points, we need to find where the function's slope is zero. In calculus terms, this means finding where the first derivative of the function, f'(x), equals zero. Stationary points are crucial in understanding the behavior of a function, as they indicate where the function changes direction, either from increasing to decreasing (a local maximum) or from decreasing to increasing (a local minimum). They can also be points of inflection where the concavity of the graph changes. So, finding these points is a fundamental step in analyzing a function.

First, let's find the derivative of f(x). Given f(x) = x³ - 6x² + 5, we'll apply the power rule for differentiation, which states that if f(x) = axⁿ, then f'(x) = naxⁿ⁻¹. This rule is the cornerstone of differentiating polynomial functions, and it's essential for finding rates of change and slopes of curves. Applying this rule term by term will give us the derivative we need.

So, differentiating x³ gives us 3x², and differentiating -6x² gives us -12x. The constant term, 5, differentiates to zero because the derivative of a constant is always zero. This is because constants do not change, and the derivative measures the rate of change. Thus, the first derivative, f'(x), represents the function's slope at any point x. Setting this derivative to zero will identify points where the function momentarily stops increasing or decreasing.

Therefore, f'(x) = 3x² - 12x. Now, we set f'(x) = 0 to find the x-values of the stationary points. This means solving the quadratic equation 3x² - 12x = 0. Solving this equation will give us the x-coordinates where the tangent to the curve is horizontal, indicating a potential maximum, minimum, or inflection point.

We can factor out a 3x from the equation, giving us 3x(x - 4) = 0. From this factored form, we can easily see that the solutions are x = 0 and x = 4. These x-values are the locations where the function's rate of change is zero, marking the potential stationary points. However, to fully define these points, we need their corresponding y-values, which we'll find in the next step by plugging these x-values back into the original function, f(x).

So, the x-coordinates of the stationary points are x = 0 and x = 4. To find the complete coordinates, we'll substitute these values back into the original function f(x) = x³ - 6x² + 5. For x = 0, we have f(0) = (0)³ - 6(0)² + 5 = 5. For x = 4, we have f(4) = (4)³ - 6(4)² + 5 = 64 - 96 + 5 = -27. Therefore, the stationary points are (0, 5) and (4, -27). These points are critical for sketching the graph of the function and understanding its behavior.

b. Determining Stationary Values of the Function f(x)

Now that we've found the stationary points, (0, 5) and (4, -27), let's determine the stationary values. The stationary values are simply the y-coordinates of the stationary points. They represent the function's value at these critical points, giving us the height of the function at each location where it momentarily flattens out. Understanding these values is essential for sketching the graph and identifying the function's range and local extrema.

From our previous calculations, we already have these values. At the stationary point (0, 5), the y-value is 5, and at the stationary point (4, -27), the y-value is -27. Therefore, the stationary values of the function f(x) are 5 and -27. These values tell us the height of the function at its critical points, where the slope is zero.

So, the stationary values are y = 5 and y = -27. These values are crucial because they help us understand the function's behavior around these points. A stationary value of 5 indicates that at x = 0, the function has a height of 5, and a stationary value of -27 indicates that at x = 4, the function dips down to -27. These values, combined with the stationary points, paint a clearer picture of the function's graph and its characteristics.

c. Determining the Type of Stationary Point for Each Point

Finally, let's determine the type of stationary point for each of the points we found: (0, 5) and (4, -27). To do this, we need to analyze the second derivative of the function, f''(x). The second derivative tells us about the concavity of the function. If f''(x) > 0, the function is concave up, indicating a local minimum. If f''(x) < 0, the function is concave down, indicating a local maximum. If f''(x) = 0, the test is inconclusive, and we might have an inflection point or need to use another method.

First, we need to find the second derivative of f(x). We already know that f'(x) = 3x² - 12x. Now, we differentiate f'(x) to find f''(x). Applying the power rule again, the derivative of 3x² is 6x, and the derivative of -12x is -12. So, f''(x) = 6x - 12. The second derivative is a linear function, and we will use it to assess the concavity at our stationary points.

Now, we'll evaluate f''(x) at each stationary point. For the point (0, 5), we have x = 0, so f''(0) = 6(0) - 12 = -12. Since f''(0) < 0, the function is concave down at x = 0, which means that (0, 5) is a local maximum. This indicates that the function reaches a peak at this point, changing from increasing to decreasing.

For the point (4, -27), we have x = 4, so f''(4) = 6(4) - 12 = 24 - 12 = 12. Since f''(4) > 0, the function is concave up at x = 4, which means that (4, -27) is a local minimum. This indicates that the function reaches a trough at this point, changing from decreasing to increasing.

Therefore, the stationary point (0, 5) is a local maximum, and the stationary point (4, -27) is a local minimum. These classifications help us sketch the graph of the function more accurately. The local maximum at (0, 5) and the local minimum at (4, -27) give us critical turning points in the function's trajectory.

In summary, by finding the first and second derivatives, we've successfully identified the stationary points and classified them as either local maxima or local minima. This process is fundamental in calculus and provides a powerful tool for analyzing functions and understanding their behavior.

Conclusion

Alright, guys, we've tackled a pretty neat calculus problem today! We started with the function f(x) = x³ - 6x² + 5 and went through the steps to find its stationary points, stationary values, and the nature of those points. Remember, stationary points are where the derivative equals zero, and they can be local maxima, local minima, or points of inflection.

We found that the stationary points were (0, 5) and (4, -27). The stationary values are the y-coordinates of these points, which are 5 and -27, respectively. By using the second derivative test, we determined that (0, 5) is a local maximum and (4, -27) is a local minimum. These details give us a solid understanding of how the function behaves.

Understanding these concepts is super useful for all sorts of applications, from physics to economics, where you need to analyze how things change and find optimal values. So, keep practicing, and you'll become a calculus whiz in no time! Great job, everyone!