Mean Value Theorem: Verification And 'c' Value Calculation

by Alex Johnson 59 views

Hey there, math enthusiasts! Today, we're diving into a classic calculus concept: the Mean Value Theorem (MVT). Specifically, we'll walk through how to check if the MVT's conditions are met and then pinpoint those special 'c' values that bring the theorem to life. We'll be working with the function f(x) = √(x + 1) over the interval [0, 3]. Get ready to brush up on your derivatives and enjoy the beauty of calculus!

Understanding the Mean Value Theorem

Before we jump into the problem, let's make sure we're all on the same page about what the Mean Value Theorem is all about. In simple terms, the MVT is a bridge between the average rate of change and the instantaneous rate of change of a function. Imagine you're driving a car. The MVT is saying that if you travel a certain distance in a certain amount of time, there has to be at least one moment in time where your instantaneous speed (what the speedometer reads) matches your average speed over the entire trip. Pretty cool, right?

Formally, the MVT states:

If a function f is:

  1. Continuous on the closed interval [a, b].
  2. Differentiable on the open interval (a, b).

Then, there exists at least one number c in the interval (a, b) such that:

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

Where:

  • f'(c) is the derivative of f evaluated at c.
  • (f(b) - f(a)) / (b - a) is the average rate of change of f over the interval [a, b].

Essentially, the theorem guarantees that there's a point c where the tangent line to the function (the instantaneous rate of change) is parallel to the secant line drawn between the endpoints of the interval (the average rate of change). Now, let's get our hands dirty with our specific function and interval. We will look at both the continuity and differentiability requirements.

Continuity

The first condition of the Mean Value Theorem requires our function to be continuous on the closed interval [a, b]. In our case, f(x) = √(x + 1) and the interval is [0, 3]. Let's think about this function. The square root function is continuous wherever it is defined. In our case, the expression inside the square root is (x + 1). This is defined for all x ≥ -1. Since our interval [0, 3] falls completely within this domain, and the function is defined, it is continuous in the interval.

Differentiability

Next, the MVT demands that f be differentiable on the open interval (a, b). Differentiability means that the derivative exists at every point in the interval. Let's find the derivative of our function.

f(x) = √(x + 1) = (x + 1)^(1/2)

Using the power rule, we get:

f'(x) = (1/2)(x + 1)^(-1/2) = 1 / (2√(x + 1)).

Now, let's consider the domain of f'(x). It is defined everywhere except where the denominator is zero, i.e., when x = -1. Since our open interval is (0, 3), and -1 is not in that interval, the derivative exists for all points in (0, 3). Thus, f(x) is differentiable on (0, 3).

Since our function is continuous on [0, 3] and differentiable on (0, 3), the Mean Value Theorem applies to our function and the interval.

Finding the Value(s) of 'c'

Now for the fun part: finding the values of c! Remember, the MVT states:

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

Let's break this down step-by-step.

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

Our interval is [0, 3], so a = 0 and b = 3. Let's find f(0) and f(3).

  • f(0) = √(0 + 1) = √1 = 1.
  • f(3) = √(3 + 1) = √4 = 2.

Step 2: Calculate the Average Rate of Change

The average rate of change is (f(b) - f(a)) / (b - a). Plugging in our values:

Average Rate of Change = (f(3) - f(0)) / (3 - 0) = (2 - 1) / 3 = 1/3.

Step 3: Set f'(c) equal to the Average Rate of Change and Solve for c

We know that f'(x) = 1 / (2√(x + 1)). Replace x with c:

f'(c) = 1 / (2√(c + 1)).

Now, set this equal to the average rate of change (1/3):

1 / (2√(c + 1)) = 1/3

Let's solve for c:

  1. Cross-multiply: 3 = 2√(c + 1).
  2. Divide by 2: 3/2 = √(c + 1).
  3. Square both sides: (3/2)^2 = c + 1.
  4. Simplify: 9/4 = c + 1.
  5. Subtract 1: c = 9/4 - 1 = 5/4.

So, c = 5/4.

Step 4: Verify that c is in the interval

We found that c = 5/4. Our interval is (0, 3). Since 5/4 = 1.25, it definitely falls within the open interval (0, 3). Therefore, the value of c that satisfies the Mean Value Theorem is c = 5/4.

Conclusion

And there you have it! We've successfully verified that the hypotheses of the Mean Value Theorem are satisfied for the function f(x) = √(x + 1) on the interval [0, 3] and found the value of c to be 5/4. This means that at the point x = 5/4, the tangent line to the curve f(x) has the same slope as the secant line connecting the points (0, f(0)) and (3, f(3)). The Mean Value Theorem is a powerful tool in calculus, connecting the concepts of average and instantaneous rates of change. Keep practicing, and you'll become a MVT master in no time!

I hope you enjoyed this journey through the Mean Value Theorem. Remember, practice makes perfect! Keep exploring and challenging yourself with calculus problems. See you in the next one!

For further exploration and deeper understanding, you can check out the resources on Khan Academy https://www.khanacademy.org/. They offer excellent tutorials and practice problems to hone your calculus skills. Happy learning!