Problem 15 - Calculus Early Transcendentals

Question

Let \(f(x)=\tan x\) (a) Show that there is no point \(c\) in the interval \((0, \pi)\) such that \(f^{\prime}(c)=0,\) even though \(f(0)=f(\pi)=0\). (b) Explain why the result in part (a) does not contradict Rolle's Theorem.

Step-by-Step Solution

Answer

The derivative of \(\tan x\) is never zero in \((0, \pi)\). Rolle's Theorem isn't violated because \(\tan x\) isn't continuous over \([0, \pi]\).

Step 1: Understand the Problem

We need to analyze the function \(f(x) = \tan x\) over the interval \((0, \pi)\), and demonstrate that there is no point \(c\) at which the derivative \(f'(c) = 0\). We'll also explore why this finding does not violate Rolle's Theorem, which typically applies under certain conditions.

Step 2: Calculate the Derivative

First, we calculate the derivative of \(f(x) = \tan x\). The derivative is given by:\[ f'(x) = \sec^2 x = \frac{1}{\cos^2 x}. \]This derivative expresses how the tangent function changes as \(x\) varies.

Step 3: Analyze the Derivative

To find where \(f'(c) = 0\), we would typically solve \(\sec^2 c = 0\) or \(\frac{1}{\cos^2 c} = 0\). However, the function \(\cos^2 c\) never equals zero for \(c\) in \((0, \pi)\), meaning \(f'(c)\) is never zero in this interval.

Step 4: Understand Function Behavior at Boundary

Observe \(f(0) = \tan(0) = 0\) and \(f(\pi) = \tan(\pi) = 0\). Even though the function equals zero at the boundaries, these points are outside the interval \((0, \pi)\), hence not satisfying the internal condition needed for Rolle's Theorem.

Step 5: Explain Rolle's Theorem Requirements

Rolle's Theorem states that for it to be applicable, a function must be continuous on \([a, b]\) and differentiable on \((a, b)\), with \(f(a) = f(b)\). However, \(f(x) = \tan x\) has vertical asymptotes and does not remain continuous over \([0, \pi]\), because \(f(x)\) approaches infinity as \(x\) approaches \(\frac{\pi}{2}\) from both sides, violating the continuity requirement.

Key Concepts

Derivative of trigonometric functions

When dealing with trigonometric functions, understanding their derivatives is crucial. For the tangent function, represented as \(f(x) = \tan x\), the derivative is calculated using the identity related to the secant function. The derivative of \(\tan x\) is \(f'(x) = \sec^2 x\). This can also be expressed as \(\frac{1}{\cos^2 x}\).

This result is important because it describes how the tangent function changes over any interval. In our exercise, we specifically analyzed this derivative between \(x = 0\) and \(x = \pi\).

• The squared secant \(\sec^2 x\) never becomes zero, since cosine squared \(\cos^2 x\) never equals zero within this interval.
• This means \(f'(c) = 0\) for \(f(x) = \tan x\) does not happen in the open interval \((0, \pi)\).

Recognizing where a function's derivative does not equate to zero helps us understand the behavior of the function graphically and ensures we are aware of where it doesn't have horizontal tangents.

Rolle's Theorem

Rolle's Theorem provides a valuable insight when studying the behavior of continuous functions on closed intervals. It asserts that if a function \(f\) is continuous on a closed interval \([a, b]\), differentiable on an open interval \((a, b)\), and \(f(a) = f(b)\), then there is at least one point \(c\) in \((a, b)\) such that \(f'(c) = 0\).

This theorem relies on three core conditions:

• The function must be continuous on the closed interval \([a, b]\).
• The function must be differentiable on \((a, b)\).
• The values at the endpoints must be equal, i.e., \(f(a) = f(b)\).

However, the tangent function over the interval \([0, \pi]\) does not meet all these criteria. Although \(f(0) = f(\pi) = 0\), it isn't continuous on the entire interval. This is because as \(x\) approaches \(\frac{\pi}{2}\), \(\tan x\) heads to infinity, creating a discontinuity. As such, the function does not satisfy the necessary conditions of Rolle’s Theorem in this case.

Tangent function

The tangent function, \(f(x) = \tan x\), is one of the basic trigonometric functions. It's a periodic function with a period of \(\pi\) and expresses the ratio of sine to cosine, i.e., \(\tan(x) = \frac{\sin(x)}{\cos(x)}\). This function has some unique properties, particularly its vertical asymptotes and points of discontinuity.

These vertical asymptotes occur at points where the cosine of \(x\) equals zero, such as \(x = \frac{\pi}{2}\), making the function undefined at these values. Over a period from \(0\) to \(\pi\), the tangent starts at zero, rises to infinity as it approaches \(\frac{\pi}{2}\), and then becomes negative as it moves past, until it returns to zero at \(\pi\).

• It is important to understand that \(\tan x\) does not remain defined through \([0, \pi]\) because of the asymptote at \(\frac{\pi}{2}\).
• This explains why the tangent function is not continuous on certain intervals, impacting the applicability of Rolle's Theorem.

In calculus and trigonometry, comprehending the nature of these discontinuities is essential for analyzing function behavior and solving related problems correctly.