Mean Value Theorem: Find 'c' For F(x) = -4x^2 + X - 2

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Hey guys! Let's dive into a classic calculus problem today: applying the Mean Value Theorem. We've got the function f(x) = -4x^2 + x - 2 defined on the interval [0, 3], and our mission is to find that special point 'c' within this interval that makes the Mean Value Theorem sing. Sounds like fun, right? Let's get started!

Understanding the Mean Value Theorem

Before we jump into the calculations, let's quickly recap what the Mean Value Theorem (MVT) is all about. In simple terms, the Mean Value Theorem states that if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one point c in the interval (a, b) such that the instantaneous rate of change (the derivative) at c is equal to the average rate of change over the interval [a, b]. This may sound a bit complicated, but trust me, it's a pretty intuitive idea once you break it down.

In mathematical notation, the Mean Value Theorem looks like this:

f'(c) = (f(b) - f(a)) / (b - a)

Where:

  • f'(c) is the derivative of the function evaluated at the point c.
  • f(b) is the function evaluated at the endpoint b.
  • f(a) is the function evaluated at the endpoint a.
  • (b - a) is the length of the interval.

Think of it this way: the left side, f'(c), represents the slope of the tangent line at the point x = c. The right side, (f(b) - f(a)) / (b - a), represents the slope of the secant line connecting the points (a, f(a)) and (b, f(b)). The Mean Value Theorem guarantees that at some point c, these two slopes are equal. This theorem provides a crucial link between the average rate of change and the instantaneous rate of change of a function, making it a cornerstone of calculus and analysis. We use this theorem to solve many problems, and understanding it well is crucial for success in calculus and beyond. It's not just a theoretical concept; it has real-world applications, from physics to economics, where understanding rates of change is paramount.

Verifying the Conditions for the Mean Value Theorem

Alright, before we go hunting for our 'c' value, we need to make sure our function f(x) = -4x^2 + x - 2 actually plays by the rules of the Mean Value Theorem on the interval [0, 3]. This means we need to check two key conditions:

  1. Continuity: Is f(x) continuous on the closed interval [0, 3]?
  2. Differentiability: Is f(x) differentiable on the open interval (0, 3)?

Our function f(x) is a polynomial, and polynomials are like the cool kids of the function world – they're continuous and differentiable everywhere! No discontinuities, no sharp corners, just smooth sailing. So, f(x) is definitely continuous on [0, 3] and differentiable on (0, 3). We've passed the first hurdle! The fact that f(x) is a polynomial greatly simplifies our task, as we don't need to worry about any potential discontinuities or points where the derivative might not exist. This is a common scenario in many calculus problems, but it's always good practice to explicitly verify these conditions to ensure we're applying the Mean Value Theorem correctly. Remember, the Mean Value Theorem is a powerful tool, but it only works if its preconditions are met. By diligently checking these conditions, we avoid making mistakes and gain a deeper understanding of the theorem's applicability.

Applying the Mean Value Theorem: Step-by-Step

Now that we've confirmed that the Mean Value Theorem applies, let's roll up our sleeves and find that elusive 'c' value. Here's how we'll do it, step-by-step:

Step 1: Calculate f(a) and f(b)

First, we need to evaluate our function at the endpoints of the interval, a = 0 and b = 3. So, let's plug those values into f(x) = -4x^2 + x - 2:

  • f(0) = -4(0)^2 + 0 - 2 = -2
  • f(3) = -4(3)^2 + 3 - 2 = -36 + 3 - 2 = -35

Piece of cake, right? We've got our function values at the endpoints.

Step 2: Calculate the average rate of change: (f(b) - f(a)) / (b - a)

Next, let's find the average rate of change of f(x) over the interval [0, 3]. This is simply the difference in function values divided by the difference in x-values:

(f(3) - f(0)) / (3 - 0) = (-35 - (-2)) / 3 = -33 / 3 = -11

So, the average rate of change is -11. This represents the slope of the secant line connecting the points (0, -2) and (3, -35) on the graph of f(x).

Step 3: Find the derivative, f'(x)

Now, we need to find the derivative of our function. Remember the power rule? It's our best friend here. Applying the power rule to f(x) = -4x^2 + x - 2, we get:

f'(x) = -8x + 1

This derivative, f'(x), gives us the instantaneous rate of change of f(x) at any point x. It's a crucial piece of the puzzle for finding our 'c' value.

Step 4: Set f'(c) equal to the average rate of change and solve for c

This is where the magic happens! According to the Mean Value Theorem, there's a point c where the instantaneous rate of change f'(c) is equal to the average rate of change we calculated earlier. So, let's set them equal and solve for c:

f'(c) = -8c + 1 = -11

Now, it's just a matter of algebra:

  • -8c = -12
  • c = -12 / -8 = 3/2 = 1.5

Ta-da! We've found our 'c' value.

Step 5: Verify that c is in the interval (0, 3)

Finally, we need to make sure that our 'c' value actually lives within the interval we're interested in, (0, 3). And guess what? 1.5 is indeed between 0 and 3. We've officially satisfied all the conditions and found the point 'c' guaranteed by the Mean Value Theorem.

Each of these steps is crucial to the application of the Mean Value Theorem. Calculating f(a) and f(b) gives us the function values at the interval's endpoints, allowing us to compute the average rate of change. Finding the derivative f'(x) is essential because it represents the instantaneous rate of change at any point. Setting f'(c) equal to the average rate of change and solving for c is the core of the theorem, as it identifies the point where the instantaneous rate of change matches the average rate. Finally, verifying that c is within the interval (0, 3) ensures that our solution is valid and consistent with the theorem's conditions.

Conclusion: The Magic of the Mean Value Theorem

So, there you have it! We successfully found the point c = 1.5 in the interval [0, 3] that satisfies the Mean Value Theorem for the function f(x) = -4x^2 + x - 2. This example beautifully illustrates the power and elegance of the Mean Value Theorem. It's not just a theoretical concept; it provides a concrete connection between the average and instantaneous rates of change, allowing us to solve problems and gain deeper insights into the behavior of functions. The Mean Value Theorem is a cornerstone of calculus, and mastering its application is crucial for any student venturing further into the world of mathematics. By understanding the conditions under which it applies and following the steps outlined above, you can confidently tackle a wide range of problems involving rates of change and function behavior. Keep practicing, and you'll become a Mean Value Theorem master in no time! Remember, the beauty of calculus lies in its ability to connect abstract concepts to real-world phenomena, and the Mean Value Theorem is a shining example of this connection. Now go forth and conquer more calculus challenges!