Problem 24 - University Calculus: Early Transcendentals

Question

Show that the functions have exactly one zero in the given interval. $$g(t)=\frac{1}{1-t}+\sqrt{1+t}-3.1, \quad(-1,1)$$

Step-by-Step Solution

Answer

The function has exactly one zero in the interval \((-1,1)\).

Step 1: Compute the function value at the endpoints

Evaluate the function \( g(t) \) at the endpoints of the interval \(-1, 1\). However, note that \( g(t) \) is not defined at \( t = 1 \) because of the term \( \frac{1}{1-t} \), which will result in division by zero. At \( t = -1 \), calculate \( g(-1) = \frac{1}{1-(-1)} + \sqrt{1+(-1)} - 3.1 = \frac{1}{2} + 0 - 3.1 = -2.6 \). The function value at \( t = 1 \) tends toward infinity, indicating a discontinuity.

Step 2: Analyze continuity and differentiability

The function \( g(t) \) is continuous and differentiable within the open interval \((-1, 1)\) except at \( t = 1 \) because it contains the terms \( \frac{1}{1-t} \) (discontinuity at \( t = 1 \)) and \( \sqrt{1+t} \). Since \( t = 1\) is not included in the interval, \( g(t) \) is continuous on the whole interval \((-1, 1)\).

Step 3: Verify the Intermediate Value Theorem

The Intermediate Value Theorem states that if \( f \) is continuous on \([a, b] \) and \( f(a) \cdot f(b) < 0 \), then \( f \) has at least one zero between \( a \) and \( b \). Since \( g(-1) = -2.6 < 0 \) and approaching infinity at \( t = 1 \), it crosses zero somewhere in \((-1, 1)\).

Step 4: Use the derivative to show a single crossing (zero)

Find the derivative \( g'(t) \) to show the function's behavior. We have \( g'(t) = \frac{1}{(1-t)^2} + \frac{1}{2\sqrt{1+t}} \). Since both terms are positive where defined in the interval \((-1, 1)\), \( g'(t) > 0 \). This implies \( g(t) \) is strictly increasing in the interval, meaning that \( g(t) \) can cross the \( t \)-axis at most once.

Key Concepts

Intermediate Value Theorem

The Intermediate Value Theorem (IVT) is a powerful tool in calculus. It deals with continuous functions and their behavior over an interval. Here’s how it works: When a continuous function goes from a negative value to a positive value over an interval, it must cross the x-axis at least once. This means the function has at least one zero, or root, within that interval.

Let’s say you have a function, and you know it is continuous on an interval \([a, b]\). If the function value at \(a\) is negative and at \(b\) it is positive, the IVT assures that the function will be zero at some point in between \(a\) and \(b\).

• Continuous functions have no breaks, jumps, or holes in the graph.
• If you plot the graph, a continuous function will allow you to draw from start to end without lifting your pencil.

The function \( g(t) \) in the exercise is continuous in the interval \((-1, 1)\), which ensures it meets the criteria needed to apply the IVT.

Continuity

Continuity is a key concept that ensures a function behaves predictably within an interval. A continuous function has no interruptions; it smoothly carries on without jumps or holes. For a function to be continuous over a certain interval, it must not only be defined at each point of the interval but also approach the same value from either side at those points.

In the given exercise, the function \({g(t) = \frac{1}{1-t} + \sqrt{1+t} - 3.1}\) is continuous on the interval \((-1, 1)\) excluding the point \(t = 1\). At \(t = 1\), the function poses a discontinuity due to the term \(\frac{1}{1-t}\) at that exact point because it leads to division by zero. However, such points are not within the interval of interest, so we focus on areas where the function remains continuous.

• On \((-1, 1)\), the function is intact and continuous.
• Continuous doesn’t mean constant; the function can vary as long as it doesn’t skip values abruptly.

This continuity is crucial for applying the Intermediate Value Theorem effectively.

Differentiability

Differentiability refers to the existence of a derivative; essentially, it tells us how smoothly a function changes. When a function is differentiable at a point, it means you can draw a tangent there, indicating a well-defined slope. Differentiable functions are a step further from continuous functions; if a function is differentiable at a point, it is indeed continuous there as well.

For \( g(t) \), the exercise focuses on the differentiability within the interval \((-1, 1)\), which is assured by the continuous nature of the function everywhere except excluding \(t = 1\). Within this stretch, the lack of sharp corners or cusps in the function plot ensures differentiability.

• A differentiable function permits the calculation of its slope at any given point in its domain (where it’s defined).
• It indicates no abrupt directional changes.

This differentiability, paired with continuity, assists in predicting function behavior - essential for understanding calculus problems deeply.

Derivative

The derivative of a function is a fundamental concept in calculus, representing the rate at which a function’s value changes. Calculating a derivative gives us the slope of the tangent to the function at any given point (where the function is differentiable).

For the function \( g(t) = \frac{1}{1-t} + \sqrt{1+t} - 3.1\), its derivative \( g'(t) = \frac{1}{(1-t)^2} + \frac{1}{2\sqrt{1+t}}\). Each component involves assessing changes:

• \(\frac{1}{(1-t)^2}\) relates to how quickly the reciprocal of the term changes.
• \(\frac{1}{2\sqrt{1+t}}\) represents changes in the square root function.

Both segments are positive within the defined interval \((-1, 1)\), indicating that \(g(t)\) is increasing. This increasing nature implies that \(g(t)\) crosses the t-axis at one point only, leading to exactly one zero in the interval, as differentiated values never decrease within this range.