Proof: $f'(c_1)+f'(c_2)=2$ With MVT

Hey guys! Let's dive into a super interesting calculus problem today. We're going to tackle a proof that involves a function $f$ that's continuous on a closed interval $[a, b]$ and differentiable on the open interval $(a, b)$. The coolest part? This function has the property that $f(a) = a$ and $f(b) = b$. Our mission, should we choose to accept it (and we totally do!), is to show that there exist two distinct points, $c_1$ and $c_2$, within the interval $(a, b)$ such that $f'(c_1) + f'(c_2) = 2$. Buckle up, it's gonna be a fun ride!

Setting the Stage: The Mean Value Theorem

Now, before we jump straight into the proof, let's arm ourselves with a powerful tool from calculus: the Mean Value Theorem (MVT). This theorem is like our trusty sidekick in many calculus adventures, and this is no exception. The MVT, in a nutshell, states that if a function $f$ 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:

$f'(c) = \frac{f(b) - f(a)}{b - a}$

Think of it this way: the MVT guarantees that there's a point $c$ where the instantaneous rate of change of the function ($f'(c)$) is equal to the average rate of change over the entire interval ($\frac{f(b) - f(a)}{b - a}$). It's like saying there's a moment on a road trip where your speedometer reading matches your average speed for the whole trip. Pretty neat, huh?

Breaking Down the Problem: Applying the MVT Cleverly

Okay, now let's get back to our original problem. We need to show that $f'(c_1) + f'(c_2) = 2$ for some distinct $c_1$ and $c_2$ in $(a, b)$. The key here is to apply the MVT not just once, but twice, and in a very strategic way. Since we want to find two points, it makes sense to split our interval $(a, b)$ into two subintervals. The most natural way to do this is to consider the midpoint of the interval, which we'll call $m = \frac{a + b}{2}$.

So, we're going to apply the MVT on the intervals $[a, m]$ and $[m, b]$ separately. This will give us two points, one in each subinterval, where the derivative of $f$ has a specific value. Let's see how this unfolds.

The First Application: Interval $[a, m]$

Let's apply the MVT on the interval $[a, m]$. Since $f$ is continuous on $[a, b]$, it's also continuous on $[a, m]$. Similarly, since $f$ is differentiable on $(a, b)$, it's also differentiable on $(a, m)$. Therefore, the MVT conditions are satisfied, and there exists a point $c_1$ in $(a, m)$ such that:

$f'(c_1) = \frac{f(m) - f(a)}{m - a}$

Remember that $m = \frac{a + b}{2}$ and $f(a) = a$. So, we can rewrite the equation as:

$f'(c_1) = \frac{f(\frac{a + b}{2}) - a}{\frac{a + b}{2} - a} = \frac{f(\frac{a + b}{2}) - a}{\frac{b - a}{2}} = \frac{2[f(\frac{a + b}{2}) - a]}{b - a}$

The Second Application: Interval $[m, b]$

Now, let's do the same thing on the interval $[m, b]$. Again, the MVT conditions are met, so there exists a point $c_2$ in $(m, b)$ such that:

$f'(c_2) = \frac{f(b) - f(m)}{b - m}$

We know that $f(b) = b$ and $m = \frac{a + b}{2}$, so we can rewrite this as:

$f'(c_2) = \frac{b - f(\frac{a + b}{2})}{b - \frac{a + b}{2}} = \frac{b - f(\frac{a + b}{2})}{\frac{b - a}{2}} = \frac{2[b - f(\frac{a + b}{2})]}{b - a}$

Putting it Together: The Grand Finale!

We've found expressions for $f'(c_1)$ and $f'(c_2)$. Now comes the moment of truth! Let's add them together and see what happens:

$f'(c_1) + f'(c_2) = \frac{2[f(\frac{a + b}{2}) - a]}{b - a} + \frac{2[b - f(\frac{a + b}{2})]}{b - a}$

Notice that both terms have a common denominator, $(b - a)$. This makes our lives much easier. Let's combine the fractions:

$f'(c_1) + f'(c_2) = \frac{2[f(\frac{a + b}{2}) - a] + 2[b - f(\frac{a + b}{2})]}{b - a}$

Now, let's distribute the 2 in the numerator:

$f'(c_1) + f'(c_2) = \frac{2f(\frac{a + b}{2}) - 2a + 2b - 2f(\frac{a + b}{2})}{b - a}$

Look closely! The terms $2f(\frac{a + b}{2})$ and $-2f(\frac{a + b}{2})$ cancel each other out. This leaves us with:

$f'(c_1) + f'(c_2) = \frac{-2a + 2b}{b - a}$

We can factor out a 2 from the numerator:

$f'(c_1) + f'(c_2) = \frac{2(b - a)}{b - a}$

And finally, we arrive at the grand conclusion:

$f'(c_1) + f'(c_2) = 2$

Boom! We've done it! We've shown that there exist distinct points $c_1$ and $c_2$ in $(a, b)$ such that $f'(c_1) + f'(c_2) = 2$. How cool is that?

Why This Matters: The Significance of the Result

Okay, so we proved a theorem. But why should we care? What's the big deal? Well, this result gives us a deeper understanding of how the derivative of a function behaves when the function has specific values at the endpoints of an interval. It tells us that the average rate of change, represented by the sum of the derivatives at two points, is constrained to a specific value (in this case, 2). This kind of result can be useful in various areas of mathematics and its applications, such as optimization problems, numerical analysis, and even physics.

Generalizing the Idea: Beyond This Specific Case

While we focused on a specific case where $f(a) = a$ and $f(b) = b$, the underlying idea of using the MVT in clever ways to prove results about derivatives is a powerful one. We can often adapt this approach to tackle similar problems with different conditions or functions. The key is to think strategically about how to break down the interval and apply the MVT to get the information we need.

Practice Makes Perfect: Try This One Out!

Now that we've conquered this problem together, how about trying a similar one on your own? This is where the real learning happens. Here's a challenge for you:

Let $g$ be continuous on $[0, 2]$ and differentiable on $(0, 2)$ with $g(0) = 0$ and $g(2) = 3$. Show that there exists a $c$ in $(0, 2)$ such that $g'(c) > 1$.

Give it a shot! Use the ideas we discussed today, and remember the power of the Mean Value Theorem. You've got this!

Conclusion: Calculus Adventures Await!

So, there you have it! We've successfully navigated a challenging calculus problem, proving a neat result about derivatives using the Mean Value Theorem. Remember, calculus is like a toolbox filled with powerful tools, and the MVT is definitely one of the handiest. Keep exploring, keep questioning, and keep having fun with math! Until next time, happy calculating, guys!