Problem 17 - Thomas Calculus

Question

Show that the functions have exactly one zero in the given interval. \(g(t)=\sqrt{t}+\sqrt{1+t}-4, \quad(0, \infty)\)

Step-by-Step Solution

Answer

The function has exactly one zero in the interval \((0, \infty)\).

Step 1: Interpret the Function and Interval

We are given the function \( g(t) = \sqrt{t} + \sqrt{1+t} - 4 \) and need to find where it has zeros, specifically within the interval \((0,\infty)\). This means finding \( t \) such that \( g(t) = 0 \).

Step 2: Examine the Behavior of the Function at the Boundaries

At \( t = 0 \), the function evaluates to \( g(0) = \sqrt{0} + \sqrt{1+0} - 4 = 1 - 4 = -3 \), indicating \( g(t) < 0 \) as \( t \to 0^+ \). As \( t \to \infty \), both \( \sqrt{t} \) and \( \sqrt{1+t} \) tend towards infinity, causing \( g(t) \to \infty \). Thus, \( g(t) \) changes from negative to positive as \( t \) increases from \( 0 \) to \( \infty \).

Step 3: Finding Intervals of Change

Through the Intermediate Value Theorem, since \( g(t) \) is continuous on the interval \((0, \infty)\), there exists at least one \( t \) in \((0, \infty)\) for which \( g(t) = 0 \) due to the change from \( g(0) < 0 \) to \( g(t) \to \infty \).

Step 4: Verify the Uniqueness of the Zero

To confirm that there is exactly one zero, consider the derivative \( g'(t) = \frac{1}{2\sqrt{t}} + \frac{1}{2\sqrt{1+t}} \), which is positive for all \( t > 0 \). This implies that \( g(t) \) is strictly increasing on \((0, \infty)\), meaning it can cross the x-axis at most once, thus having exactly one zero.

Key Concepts

Continuous Functions

A continuous function is one that shows no breaks, jumps, or interruptions in its behavior across its domain. Such functions smoothly connect point-to-point without any gaps.
In mathematics, this is formalized by saying that for any point within the domain of the function, as you approach that point from any direction, the function value approaches the same limit. This property is crucial for many mathematical theorems.

Why is continuity important in the Intermediate Value Theorem? Simply put, the Intermediate Value Theorem relies heavily on the continuity of functions. Let's break it down:

• If a function, like our given function, is continuous on a particular interval, then any change from negative to positive values implies there must be a point or points within that interval where the function equals zero.
• This assures us of the existence of a solution without precisely calculating where it occurs.

For the function \( g(t) = \sqrt{t} + \sqrt{1+t} - 4 \) over the interval \((0,\infty)\), the continuous nature of the square root and linear operations ensures no disruptions.
Thus, it is safe to use the Intermediate Value Theorem to deduce the existence of zeros.

Function Behavior Analysis

Understanding the behavior of a function is critical for solving problems like finding zeros. It involves checking how a function acts as input values change. For example, by testing function values at certain points or examining limits as values approach certain boundaries.

For \( g(t) \), our primary points of interest are the boundaries of our interval, zero, and infinity:

• At \( t = 0 \), by evaluating \( g(0) = \sqrt{0} + \sqrt{1+0} - 4 = -3 \), we find the function starts negative.
• Considering \( t \to \infty \), both the terms \( \sqrt{t} \) and \( \sqrt{1+t} \) grow without bound, pushing \( g(t) \) towards positive infinity.

This behavior preview helps forecast potential intersections with the x-axis. Because \( g(t) \) transitions from negative to positive, we suspect it must cross the x-axis, showing a zero exists.

Derivatives

Derivatives provide an insightful lens into how a function is changing at any point within its domain. By deriving \( g(t) \), we capture the rate of change and the nature of the function's slope.

For our function, the derivative \( g'(t) = \frac{1}{2\sqrt{t}} + \frac{1}{2\sqrt{1+t}} \) reveals a few important aspects:

• This derivative's positivity for all \( t > 0 \), as both terms are always positive where they are defined, tells us that the function is always increasing as \( t \) increases.
• The increase is dictated by the magnitude of these terms, appending continuously increasing behavior.

By understanding this derivative, we confirm the function does not dip down again after crossing the x-axis, ensuring the unique nature of the zero.

Strictly Increasing Functions

A strictly increasing function continually grows, never faltering or showing decrease for any increase in the input variable. Such behavior implies a clear path for showing the uniqueness of intersections with the x-axis.

For \( g(t) \), since \( g'(t) > 0 \) everywhere in its domain, \( g(t) \) is strictly increasing:

• This prohibits any backtracking or multiple crossings of the x-axis, solidifying the presence of at most one zero.
• A strictly increasing function rising from negative to positive over an interval means a singular solution with no possibility for multiple answers, as it can't double back.
• This characteristic eases worries about additional hidden zeros.

Thus, knowing \( g(t) \) doesn't have dips ensures one cross at one point, confirming exactly one zero in \((0,\infty)\). This makes strictly increasing behavior a powerful tool in verifying unique solutions.