Question

A function \(f(x)\) and interval \([a, b]\) are given. Check if the Mean Value Theorem can be applied to \(f\) on \([a, b] ;\) if so, find a value \(c\) in \([a, b]\) guaranteed by the Mean Value Theorem. \(f(x)=\tan x\) on \([-\pi / 4, \pi / 4]\)

Short answer

The MVT can be applied; the value of \(c\) satisfies \(\cos c = \frac{\sqrt{\pi}}{2}\).

Step-by-step solution

Verify Continuity

The Mean Value Theorem (MVT) requires that the function be continuous on the closed interval \([a, b]\). The function \(f(x) = \tan x\) is continuous on \([-\pi / 4, \pi / 4]\) because there are no points of discontinuity within this interval. Thus, \(f(x)\) is continuous on \([a, b]\).

Verify Differentiability

The MVT also requires that the function be differentiable on the open interval \((a, b)\). The derivative of \(f(x) = \tan x\) is \(f'(x) = \sec^2 x\). Since \sec x\ is undefined where \(x = \pi/2 + k\pi\) for any integer \(k\), and \(\pi/2\) is not in \((-\pi / 4, \pi / 4)\), \(f(x)\) is differentiable on \((a, b)\).

Apply the Mean Value Theorem

The MVT states that there exists at least one value \(c\) in \((a, b)\) such that \(f'(c) = \frac{f(b) - f(a)}{b - a}\). Calculate \(f(b) - f(a)\) and \(b - a\). Here, \(f(\pi / 4) = 1\) and \(f(-\pi / 4) = -1\), so \(f(\pi / 4) - f(-\pi / 4) = 1 - (-1) = 2\). The interval length \(b - a\) is \(\pi / 4 - (-\pi / 4) = \pi / 2\).

Solve for c

We have \(f'(c) = \frac{2}{\pi/2}\). Simplify to find \(f'(c) = \frac{4}{\pi}\). Since \(f'(x) = \sec^2 x\), set \(\sec^2 c = \frac{4}{\pi}\). Solving for \(c\), we have \(\sec c = \sqrt{\frac{4}{\pi}}\). Thus, find \(c\) by solving \(\cos c = \frac{\sqrt{\pi}}{2}\). The value of \(c\) will be in \((- \pi / 4, \pi / 4)\).

Determine the Exact Value

The exact value of \(c\) where \(\cos c = \frac{\sqrt{\pi}}{2}\) must be determined through calculation or approximation, and should satisfy the condition \(-\pi / 4 < c < \pi / 4\). Verify that this value fulfills all the conditions of the problem.

Key concepts

Understanding Continuity

In calculus, continuity is a fundamental concept. A function is said to be continuous on a closed interval \([a, b]\) if there are no breaks, jumps, or holes in its graph on that interval. Imagine drawing the graph without lifting your pencil.
For the Mean Value Theorem (MVT) to apply, the function in question must be continuous over \([a, b]\).
In our exercise, we considered \(f(x) = \tan x\) over the interval \([-\pi/4, \pi/4]\). Tan functions can have points of discontinuity where the function goes to infinity, but on our given interval, \(\tan x\) is smooth and continuous without interruptions.
This means that there's no point within \([-\pi/4, \pi/4]\) where the function is undefined or jumps, making it a perfect candidate for checking if MVT can be applied.

Exploring Differentiability

Differentiability refers to the ability for a function to have a derivative at any given point in its domain. A function is differentiable on an open interval \( (a, b) \) if it has a derivative at every point in that interval.
The Mean Value Theorem also requires the function to be differentiable on this open interval. This implies the need for the function to be smooth and not have sharp corners or cusps.
In our task, \(f(x) = \tan x\) was checked for differentiability in the interval \( (-\pi/4, \pi/4) \). The derivative is \(f'(x) = \sec^2 x\), which is defined and smooth wherever \(\tan x\) is continuous, avoiding any points where \(\cos x = 0\). None of those problem points lie in our interval. Thus, you can safely say \(f(x)\) is differentiable in the required interval.

Connecting Trigonometric Functions

Trigonometric functions like \(\tan x\), \(\cos x\), and \(\sec x\) surface frequently in calculus problems. \(\tan x\) is a periodic function derived from \(\sin x / \cos x\). Understanding this periodicity and the relationship among these functions is vital in solving calculus problems that involve the Mean Value Theorem.
The secant function \(\sec x\) is directly related to \(\cos x\) such that \(\sec x = 1/\cos x\). This relationship is useful for determining differentiability, as seen with the derivative \(\sec^2 x\) in our problem.
By recognizing how these functions operate within intervals, especially their undefined points, it becomes easier to analyze their behavior over the given domain and ensure they meet the conditions of the MVT.

Calculus Fundamentals

Calculus is the mathematics of change and is crucial for understanding dynamics in various fields. At its core, it handles concepts like limits, derivatives, and integrals.
The Mean Value Theorem is a staple of calculus. It essentially states that for a continuous and differentiable function on an interval, there is at least one point where the instantaneous rate of change (derivative) equals the average rate of change over the interval.
In other words, the theorem provides a formal way to find a specific point within an interval that represents the overall behavior of the function from \( a \) to \( b \). It's a helpful tool to predict future values, optimize solutions, or simply demonstrate how a function behaves over a given range.
Our problem used these fundamentals to identify a specific value \((c)\) within \([-\pi/4, \pi/4]\), proving the theorem by showing such a point exists where the conditions are met.