Mean Value Theorem: Finding Average Slope And Value 'c'
Hey guys! Let's dive into a classic calculus problem involving the Mean Value Theorem. We've got a function, a specific interval, and a mission to find the average slope and a special point 'c' where the instantaneous slope matches that average. Buckle up, it's gonna be a fun ride!
Understanding the Problem
Before we jump into calculations, let's make sure we're all on the same page. We're given the function f(x) = 2x³ - 15x² - 36x + 5, which is a polynomial function – nice and smooth, no surprises there. We're looking at this function specifically on the interval [-4, 7], meaning we care about its behavior between x = -4 and x = 7, inclusive.
Our first task is to find the average slope of this function over this interval. Think of it like this: if you were to draw a straight line connecting the points on the graph of f(x) at x = -4 and x = 7, what would the slope of that line be? That's our average slope. And this average slope is not just some random calculation; it gives us a bird’s-eye view of how the function is changing on average across the interval. Remember that average slope sets the stage for understanding the overall trend of the function's change.
But here's where it gets interesting. The Mean Value Theorem (MVT) comes into play. This theorem guarantees that somewhere within the open interval (-4, 7), there's a point 'c' where the instantaneous rate of change of the function – its derivative, f'(x) – is exactly equal to this average slope we just calculated. So, we're not just finding the average slope; we're also tasked with pinpointing this magical point 'c' where the function's instantaneous behavior mirrors its average behavior. This part is crucial because the Mean Value Theorem doesn't just assert the existence of such a point; it underscores a fundamental relationship between the average and instantaneous rates of change of a function. So, we are tasked with the challenge of pinpointing this magical point 'c' where the function's instantaneous behavior mirrors its average behavior. The power of the Mean Value Theorem lies in its ability to bridge the macroscopic view (average change) with the microscopic view (instantaneous change), revealing a deeper understanding of function behavior.
In essence, we're going on a mathematical treasure hunt. The treasure? The value of 'c'. The map? The Mean Value Theorem. Let's get started!
Calculating the Average Slope
Okay, let's get our hands dirty with some calculations. Remember, the average slope of a function f(x) over an interval [a, b] is simply the change in y (the function's value) divided by the change in x. Mathematically, we express this as:
Average Slope = (f(b) - f(a)) / (b - a)
In our case, a = -4 and b = 7. So, the first step is to figure out the function's value at these two points. Let's plug them into our function, f(x) = 2x³ - 15x² - 36x + 5:
- f(-4) = 2(-4)³ - 15(-4)² - 36(-4) + 5 = -128 - 240 + 144 + 5 = -219
- f(7) = 2(7)³ - 15(7)² - 36(7) + 5 = 686 - 735 - 252 + 5 = -296
Now we have the y-values corresponding to our x-values. We can now determine the average slope. Let’s substitute these values into our formula:
Average Slope = (-296 - (-219)) / (7 - (-4)) = (-296 + 219) / (7 + 4) = -77 / 11 = -7
So, the average slope of our function f(x) over the interval [-4, 7] is -7. This means that, on average, for every 1 unit increase in x within this interval, the function's value decreases by 7 units. Now, this average slope serves as the benchmark against which we'll find a specific point where the instantaneous rate of change matches this overall trend. Imagine a roller coaster track, our function’s graph, and the average slope is like a straight line connecting the starting and ending points of a section of the track. Somewhere along that section, the coaster's instantaneous slope will match the slope of that straight line, giving us a moment-to-moment snapshot that mirrors the overall change.
Now that we've found the average slope, we're one step closer to finding that special 'c' value guaranteed by the Mean Value Theorem. Let's move on to the next step: finding the derivative of our function.
Finding the Derivative f'(x)
To find the point 'c' where the instantaneous slope matches our calculated average slope, we need to find the derivative of our function, f(x). Remember, the derivative, f'(x), gives us the instantaneous rate of change of the function at any given point x. Our function is f(x) = 2x³ - 15x² - 36x + 5. Let's use the power rule (which states d/dx(x^n) = nx^(n-1)) to find its derivative.
- The derivative of 2x³ is 2 * 3x² = 6x²
- The derivative of -15x² is -15 * 2x = -30x
- The derivative of -36x is -36
- The derivative of the constant 5 is 0
Putting it all together, we get:
f'(x) = 6x² - 30x - 36
This, guys, is our derivative function! It tells us the slope of the tangent line to the graph of f(x) at any point x. Now we have the tool we need to find the value(s) of x where the instantaneous slope, f'(x), is equal to our calculated average slope of -7. Our next step involves setting f'(x) equal to -7 and solving for x, which will lead us to our desired 'c' value(s).
By determining the derivative, we've essentially equipped ourselves with a powerful lens that allows us to zoom in on the instantaneous behavior of the function at any point. The derivative acts as a slope-detecting device, revealing how steeply the function is changing at each infinitesimal step along its path. This contrasts with the average slope, which gives us a broad overview of the function’s change across the entire interval.
Now, with both the average slope and the derivative function in hand, we are perfectly positioned to find the magical point 'c' where these two perspectives converge – where the instantaneous change perfectly mirrors the overall average change.
Applying the Mean Value Theorem
The Mean Value Theorem (MVT) 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 (a, b) such that:
f'(c) = (f(b) - f(a)) / (b - a)
In simpler terms, there's a point 'c' where the instantaneous slope (f'(c)) equals the average slope over the interval. We've already calculated the average slope to be -7 and found the derivative f'(x) = 6x² - 30x - 36. Now, we need to find the value(s) of 'c' that satisfy the MVT. To do this, we set f'(c) equal to the average slope and solve for c:
6c² - 30c - 36 = -7
Let's rearrange this into a quadratic equation:
6c² - 30c - 29 = 0
Now, we need to solve this quadratic equation for c. Since this doesn't factor easily, we'll use the quadratic formula:
c = (-b ± √(b² - 4ac)) / 2a
Where a = 6, b = -30, and c = -29 (the constant term in our quadratic equation). Let's plug these values in:
c = (30 ± √((-30)² - 4 * 6 * -29)) / (2 * 6) c = (30 ± √(900 + 696)) / 12 c = (30 ± √1596) / 12 c = (30 ± 2√399) / 12 c = (15 ± √399) / 6
This gives us two possible values for c:
- c₁ = (15 + √399) / 6 ≈ 5.909
- c₂ = (15 - √399) / 6 ≈ -0.909
Now, we need to check if these values fall within our open interval (-4, 7). Both c₁ ≈ 5.909 and c₂ ≈ -0.909 do indeed lie within this interval. So, we have found two points, not just one, where the instantaneous slope of the function equals the average slope over the interval [-4, 7]. This is perfectly fine, as the MVT guarantees at least one such point, but there can be more.
The quadratic formula acts as a powerful algebraic compass, guiding us through the landscape of the derivative function to precisely locate the points where the instantaneous slope aligns with the average slope. It’s a testament to the elegance of mathematical tools, allowing us to dissect and understand the intricate behavior of functions. So, in our context, the MVT acts as a bridge, connecting the macroscopic view of average slope across an interval with the microscopic view of instantaneous change at specific points, and the quadratic formula is one of the tools that helps us to pinpoint these crucial connection points.
Conclusion
Alright, guys, we've successfully navigated this Mean Value Theorem problem! We started with the function f(x) = 2x³ - 15x² - 36x + 5 and the interval [-4, 7]. Our goal was to find the average slope of the function over this interval and then find the value(s) of 'c' within the interval where the instantaneous slope, f'(c), equals that average slope, as guaranteed by the Mean Value Theorem.
We calculated the average slope to be -7. Then, we found the derivative of our function, f'(x) = 6x² - 30x - 36. By setting f'(c) equal to the average slope and solving the resulting quadratic equation, we found two values of c: approximately 5.909 and -0.909. Both of these values lie within our open interval (-4, 7), confirming the Mean Value Theorem.
So, in a nutshell, we've not only verified the Mean Value Theorem for this specific function and interval, but we've also gained a deeper understanding of how the theorem works. We've seen how the average slope provides an overall picture of the function's change, and how the derivative allows us to zoom in and find points where the instantaneous rate of change matches that average. The Mean Value Theorem is a cornerstone of calculus, connecting average behavior with instantaneous behavior, and we've just seen it in action!
