Problem 54 - A Textbook of Differential Calculus

Question

Number of points of discontinuity of \(f(x)=\tan ^{2} x-\sec ^{2} x\) in \((0,2 \pi)\) is ..........

Step-by-Step Solution

Answer

The function has 2 points of discontinuity.

Step 1: Identify Function Components

The given function is \(f(x)= \tan^2 x - \sec^2 x\). Recall the trigonometric identities: \(\tan x = \frac{\sin x}{\cos x}\) and \(\sec x = \frac{1}{\cos x}\). The function involves both tangent and secant functions.

Step 2: Rewrite the Function

Using trigonometric identities, rewrite \(f(x)\):\[ f(x) = \left( \frac{\sin x}{\cos x} \right)^2 - \left( \frac{1}{\cos x} \right)^2 = \frac{\sin^2 x}{\cos^2 x} - \frac{1}{\cos^2 x} \]Simplify this to:\[ f(x) = \frac{\sin^2 x - 1}{\cos^2 x} \]

Step 3: Simplify Using Pythagorean Identity

We know that the Pythagorean identity is \(\sin^2 x + \cos^2 x = 1\). Thus, \(\sin^2 x - 1 = -\cos^2 x\). Substitute this into the expression:\[ f(x) = \frac{-\cos^2 x}{\cos^2 x} = -1 \]

Step 4: Determine Discontinuity Points

Since \(f(x) = -1\) is a constant function, it is continuous everywhere in its domain, except where the original functions \(\tan x\) and \(\sec x\) are not defined. Recall that these functions are undefined where \(\cos x = 0\). Within the interval \((0, 2\pi)\), \(\cos x = 0\) at \(x = \frac{\pi}{2}\) and \(x = \frac{3\pi}{2}\). These are the points of discontinuity.

Step 5: Count Discontinuity Points

Thus, the function has points of discontinuity at \(x = \frac{\pi}{2} \) and \(x = \frac{3\pi}{2} \) within \( (0, 2\pi) \). Count these points to find there are two points of discontinuity.

Key Concepts

Trigonometric Identities

Understanding trigonometric identities is crucial when working with functions like \( \tan x \) and \( \sec x \). These identities allow us to rewrite and simplify expressions involving trigonometric functions. At the heart of this exercise, we used two primary trigonometric identities:

• \( \tan x = \frac{\sin x}{\cos x} \)
• \( \sec x = \frac{1}{\cos x} \)

By recognizing these relationships, we can transform complex trigonometric expressions into simpler forms. Additionally, the Pythagorean identity, \( \sin^2 x + \cos^2 x = 1 \), helped to further simplify our function to a constant. This identity embodies how the sine and cosine squared values add up to one, which is essential. In this case, knowing that \( \sin^2 x - 1 = -\cos^2 x \) helped reduce the expression effectively. Understanding these identities can make or break your ability to solve trigonometric problems efficiently.

Discontinuous Functions

In mathematics, a function is said to be discontinuous at a point if it is not continuous there. For the function \( f(x)= \tan^2 x - \sec^2 x \), the given step-by-step solution illustrates that the function simplifies to a constant \( -1 \). Although it’s a constant function, discontinuities are introduced by the nature of tangent and secant functions from which it is derived.Tangent and secant functions are discontinuous where \( \cos x = 0 \), since division by zero is undefined. In trigonometry, this typically occurs when the angle makes the cosine function reach zero. Hence, irrespective of the simplification, these inherited points affect the continuity of \( f(x) \).It's essential to account for these points, especially within specific intervals, because they indicate where the function cannot be evaluated traditionally. Understanding where these discontinuities occur helps prevent potential pitfalls when evaluating a function over an interval.

Interval Analysis

Interval analysis plays a significant role in identifying where certain properties of functions, like continuity and discontinuity, occur. When analyzing the interval \((0, 2\pi)\), we examine \( f(x) = \tan^2 x - \sec^2 x \) to determine any discontinuities.Within this specific interval, \( \cos x \) is zero at \( x = \frac{\pi}{2} \) and \( x = \frac{3\pi}{2} \). Because \( \cos x \) appears in the denominator of the original tangent and secant functions, these points lead to division by zero, making them discontinuities in the interval.Understanding this helps in

• Predicting where a function might "break" or be undefined.
• Avoiding undefined points in calculations or graphing efforts.
• Counting exact points where such occurrences happen.

The interval from \( 0 \) to \( 2\pi \) contains two discontinuities due to these specific values of \( x \), highlighting the vital role interval analysis plays in function evaluation.