Problem 36 - Essential Calculus

Question

Determine whether the Mean Value Theorem can be applied to \(f\) on the closed interval \([a, b] .\) If the Mean Value Theorem can be applied, find all values of \(c\) in the open interval \((a, b)\) such that \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\). $$ f(x)=\arctan (1-x),[0,1] $$

Step-by-Step Solution

Answer

Yes, the Mean Value Theorem can be applied to the function \(f(x)=\arctan (1-x)\) on the interval [0,1]. The value of \(c\) on the interval (0,1) that fulfills the Mean Value Theorem is \(c = 1-\frac{1}{\sqrt{\pi -1}}\)

Step 1: Check the Criteria for Mean Value Theorem

The function \(f(x)=\arctan (1-x)\) is continuous on the closed interval [0,1] and differentiable on the open interval (0,1). Therefore, the Mean Value Theorem can be applied.

Step 2: Find the Derivative of the Function

The derivative of \(f(x)=\arctan (1-x)\) can be found using the chain rule. The derivative is: \(f'(x)=-\frac{1}{(1-x)^2+1}\)

Step 3: Calculate the Average Rate of Change

The average rate of change of \(f(x)=\arctan (1-x)\) on the interval [0,1] is: \(\frac{f(1)-f(0)}{1-0}=\frac{\arctan(0) - \arctan(1)}{1} =-\frac{\pi}{4}\)

Step 4: Equate the Derivative to the Average Rate of Change

To find the values of \(c\) that meet the condition of the Mean Value Theorem, we equate the derivative to the average rate of change: \(-\frac{1}{(1-c)^2+1} = -\frac{\pi}{4}\)

Step 5: Solve for \(c\)

To solve for \(c\), simplify the previous equation and use algebra to isolate \(c\). The solution is \(c = 1-\frac{1}{\sqrt{\pi -1}}\)

Key Concepts

Continuous Function

For the Mean Value Theorem (MVT) to be applied to a function, it is crucial that the function is continuous over the closed interval \[a, b\]. A continuous function is one where you can draw its entire graph without lifting your pencil—there are no breaks or holes. In simpler terms, every point along this interval is defined, and the function behaves predictably in this range.
This requirement ensures the function doesn't have any instantaneous jumps or undefined points, thereby allowing MVT to examine rates of change.

• Continuous functions can have values that change gradually over specific intervals.
• In the case of the exercise, \( ext{arctan}(1-x)\) is continuous on \[0,1\], satisfying the first condition for MVT.

Understanding how continuity relates to the Mean Value Theorem allows you to identify functions and intervals where MVT can be put into practice.

Differentiable Function

Differentiability is the second requirement for applying the Mean Value Theorem. A function is differentiable on an interval if it has a derivative at each point within that interval. Simply put, the function must be smooth without any sharp bends or cusps.
If a function is differentiable, it means you can find its derivative at every point in the interval, which represents the rate of change at that point. This smoothness condition is essential for ensuring a well-defined slope.

• In our problem, the function \( ext{arctan}(1-x)\) is differentiable over the open interval \(0, 1\).
• The derivative, calculated using the chain rule, supports finding precise changes between points.

Differentiability ensures that MVT can find specific points where the instantaneous rate of change equals the average rate of change over the interval.

Average Rate of Change

The average rate of change of a function over an interval \[a, b\] represents how much the function value changes, on average, per unit of change in \(x\). It is given by the formula \[\frac{f(b)-f(a)}{b-a}\].
This concept is similar to finding the slope of a line connecting two points on the graph of a function. For the Mean Value Theorem, this average rate is compared to instantaneous rates to find the specific \(c\) that makes them equal.

• In the exercise, compute \[\frac{f(1)-f(0)}{1-0}\] for function \( ext{arctan}(1-x)\), giving \(-\frac{\pi}{4}\).
• This value represents the slope of the secant line over \[0,1\].

The importance of average rate of change lies in its role in connecting to derivatives, thus facilitating the application of the Mean Value Theorem.

Chain Rule

The chain rule is a fundamental concept used to differentiate composite functions. When a function is composed of two or more functions, the chain rule helps find its derivative.
Breaking this down, if you have a function \(g(x)\) inside another function \(f(x)\), the derivative is determined by differentiating the outer function \(f\), and multiplying by the derivative of the inner function \(g\).

• For \( ext{arctan}(1-x)\), we use the chain rule where \(u = 1-x\) and differentiate \(\text{arctan}(u)\).
• The result, \(-\frac{1}{(1-x)^2+1}\), shows how changes in \(x\) affect \(f(x)\).

By efficiently applying the chain rule, we derive precise expressions needed to solve for the instantaneous rate of change and facilitate the comparison with the average rate for the Mean Value Theorem.