Problem 46 - Calculus of a Single Variable

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 \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\). If the Mean Value Theorem cannot be applied, explain why not. \(f(x)=\cos x+\tan x, \quad[0, \pi]\)

Step-by-Step Solution

Answer

Yes, Mean Value Theorem can be applied to this function. The \(c\) values can be found by solving the equation: -\(\sin c+ \sec^2c=\frac{f(\pi)-f(0)}{\pi-0}\)

Step 1: Check Mean Value Theorem conditions

The Mean Value Theorem can be applied for a function if it is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\).The function \(f(x)=\cos x+\tan x\) is continuous and differentiable in the given interval \([0, \pi]\) since both cosine and tangent are continuous and differentiable on said interval. So, the Mean Value Theorem applies.

Step 2: Application of Mean Value Theorem

Apply the Mean Value Theorem which states that there is at least one point \(c\) in the open interval \((a, b)\) such that \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\)

Step 3: Find Prime Function and compute averages

Get the derivative of the given function, which gives \(f^{\prime}(x)= -\sin x+ \sec^2x\). Then calculate \(\frac{f(b)-f(a)}{b-a}\). In our case, \(a=0\) and \(b=\pi\). So we plug the values into the equation to obtain \(\frac{f(\pi)-f(0)}{\pi-0}\)

Step 4: Find Values of \(c\)

To find the values of `c` that satisfy the Mean Value Theorem, equate the average rate of change computed in step 3 to the derivative obtained in Step 3 and solve for \(c\): -\(\sin c+ \sec^2c=\frac{f(\pi)-f(0)}{\pi-0}\)

Key Concepts

Continuous Function

A continuous function is a type of function where small changes in the input result in small changes in the output. Put simply, the graph of a continuous function can be drawn without lifting your pen from the paper.
To understand this better, let's consider the function given in the exercise:

• The function is \( f(x) = \cos x + \tan x \).
• Both \( \cos x \) and \( \tan x \) are continuous functions on the interval \([0, \pi]\) except at the points where \( \tan x \) has vertical asymptotes.

However, within the specific interval \([0, \pi]\) considered in this exercise, \( \tan x \) does not have any discontinuities. Thus, \( f(x) = \cos x + \tan x \) is continuous on \([0, \pi]\). This is crucial because continuity on a closed interval is one of the necessary conditions for applying the Mean Value Theorem.

Differentiable Function

Differentiability means a function has a defined derivative at each point in its domain. A differentiable function provides us with a way to understand rates of change.
In our exercise, the function \( f(x) = \cos x + \tan x \) needs to be differentiable over the interval \((0, \pi)\).
To check for differentiability:

• The derivative of \( \cos x \) is \(-\sin x\).
• The derivative of \( \tan x \) is \( \sec^2 x \).
• Thus, \( f'(x) = -\sin x + \sec^2 x \).

Both \(-\sin x\) and \(\sec^2 x\) are well-defined in \((0, \pi)\), which means \( f(x) \) meets the differentiability requirement within this interval. This makes the Mean Value Theorem applicable in the case of our function.

Trigonometric Functions

Trigonometric functions like \(\cos x\) and \(\tan x\) are essential to understanding many problems in calculus. They behave in unique ways that influence their continuity and differentiability.
Key aspects include:

• \(\cos x\) is periodic with a range from -1 to 1, smooth and continuous everywhere.
• \(\tan x\) has discontinuities (vertical asymptotes) at odd multiples of \(\frac{\pi}{2}\), affecting its continuity, but not within \([0, \pi]\).

In the given exercise:

• The combination of \(\cos x\) and \(\tan x\) within the interval \([0, \pi]\) provides a function that is continuous and differentiable.

Understanding the properties of these trigonometric functions is important in evaluating when and where the Mean Value Theorem can be applied.

Calculus Problem Solving

Solving calculus problems involves understanding the logic and tools used in calculus. The Mean Value Theorem is a fundamental concept in calculus problem solving.
When solving these types of problems:

• First, verify that the function is continuous on the closed interval and differentiable on the open interval.
• Next, compute the derivative of the function to find the rate of change.
• Then, calculate the average rate of change over the given interval.
• Finally, equate the derivative to the average rate to find the specific value of \(c\) such that \(f'(c) = \frac{f(b) - f(a)}{b-a}\).

By understanding and applying these steps properly, you'll be able to effectively tackle calculus problems involving the Mean Value Theorem, like the one studied here.