Calculating Derivatives: Find F'(2) For F(x) = X³ - 2x + 5
Hey everyone! Today, we're diving into the world of calculus to understand derivatives and how to calculate them. Specifically, we're going to tackle a problem where we need to find the value of the derivative of a polynomial function at a specific point. So, grab your thinking caps, and let's get started!
What is a Derivative?
Before we jump into the problem, let's quickly recap what a derivative actually is. In simple terms, the derivative of a function f(x) at a point x = a, denoted as f'(a), gives us the instantaneous rate of change of the function at that point. Think of it as the slope of the tangent line to the curve of the function at x = a. This concept is super useful in various fields, from physics (calculating velocity and acceleration) to economics (analyzing marginal cost and revenue).
The derivative, f'(a), represents the instantaneous rate of change of y = f(x) with respect to x when x = a. In simpler terms, it tells us how much the function's output is changing at a very specific point. It's like zooming in infinitely close to a curve and finding the slope of the line that just touches the curve at that point. Understanding this concept is crucial because derivatives have wide applications across various fields, including physics, engineering, economics, and computer science. For instance, in physics, the derivative of a position function gives you the velocity, and the derivative of the velocity gives you acceleration. In economics, derivatives are used to analyze marginal costs and revenues. Therefore, mastering derivatives is a fundamental step in understanding more complex mathematical and scientific concepts. The derivative, at its core, is a measure of sensitivity to change. It allows us to quantify how a function responds to tiny alterations in its input. This sensitivity analysis is invaluable in optimization problems, where we seek to find the maximum or minimum values of a function. For example, businesses use derivatives to optimize production levels and minimize costs. Engineers use them to design efficient structures and systems. The power of the derivative lies in its ability to provide precise information about the behavior of a function at any given point, making it an indispensable tool in mathematical analysis.
The Problem: Finding f'(2)
We are given the polynomial function f(x) = x³ - 2x + 5, and our mission is to find the value of the derivative of this function at the point x = 2, which is written as f'(2). To do this, we'll follow these steps:
- Find the derivative of the function, f'(x).
- Substitute x = 2 into the derivative, f'(x), to find f'(2).
Let's get to it!
Step 1: Finding the Derivative f'(x)
To find the derivative f'(x), we'll use the power rule. The power rule states that if f(x) = xⁿ, then f'(x) = nxⁿ⁻¹. We'll apply this rule to each term in our function:
- For the term x³, the derivative is 3x².
- For the term -2x, the derivative is -2 (since the power of x is 1).
- For the constant term 5, the derivative is 0 (the derivative of a constant is always zero).
Combining these results, we get:
f'(x) = 3x² - 2 + 0 = 3x² - 2
So, the derivative of our function f(x) = x³ - 2x + 5 is f'(x) = 3x² - 2.
Step 2: Evaluating f'(2)
Now that we have the derivative f'(x), we can find f'(2) by substituting x = 2 into the expression:
f'(2) = 3(2)² - 2
Let's simplify this:
f'(2) = 3(4) - 2 = 12 - 2 = 10
Therefore, the value of the derivative of the function f(x) = x³ - 2x + 5 at the point x = 2 is f'(2) = 10.
The derivative, f'(x), of a function f(x) is another function that gives the slope of the tangent line to the curve of f(x) at any point x. It's a fundamental concept in calculus and is used to analyze the behavior of functions. In our case, we start with f(x) = x³ - 2x + 5. To find f'(x), we apply the power rule, which states that if f(x) = axⁿ, then f'(x) = naxⁿ⁻¹. Applying this rule to each term in our function:
- The derivative of x³ is 3x².
- The derivative of -2x is -2.
- The derivative of 5 (a constant) is 0.
Combining these, we get f'(x) = 3x² - 2. This new function, f'(x), tells us the slope of the original function f(x) at any point x. Now that we have f'(x), we can find f'(2) by simply substituting x = 2 into f'(x). This gives us f'(2) = 3(2)² - 2 = 3(4) - 2 = 12 - 2 = 10. So, f'(2) = 10. This means that at the point x = 2, the slope of the tangent line to the curve of f(x) = x³ - 2x + 5 is 10. In other words, the function is increasing at a rate of 10 units for every one unit increase in x at that point. This is a powerful piece of information that can be used to understand the local behavior of the function around x = 2. For example, we can use this information to approximate the value of f(x) at points close to x = 2 using the tangent line approximation.
Wrapping Up
So, there you have it! We've successfully found that f'(2) = 10 for the function f(x) = x³ - 2x + 5. This means that at the point x = 2, the function is changing at a rate of 10. Derivatives are a fundamental concept in calculus, and understanding how to calculate them is essential for solving various problems in mathematics, science, and engineering.
Keep practicing, and you'll become a derivative master in no time! Good luck, and happy calculating!
To summarize, finding f'(2) involves two key steps: first, determining the derivative f'(x) of the given function f(x), and second, evaluating f'(x) at x = 2. In our example, f(x) = x³ - 2x + 5. The derivative f'(x) is found using the power rule, resulting in f'(x) = 3x² - 2. Then, substituting x = 2 into f'(x) gives us f'(2) = 3(2)² - 2 = 10. This result indicates that at x = 2, the function f(x) is changing at a rate of 10. This process showcases the power of derivatives in analyzing the behavior of functions at specific points. Understanding and applying these steps correctly is crucial for solving more complex problems in calculus and its applications. Remember to practice these steps with different functions to build your proficiency and confidence. The ability to find and interpret derivatives is a valuable skill that will serve you well in various fields of study and professional endeavors.