Problem 21 - Thomas Calculus in SI Units

Question

Show that the functions that 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 crosses the x-axis exactly once since it changes from negative to positive and is strictly increasing in the interval.

Step 1: Analyze the given function

We are given the function \( g(t) = \sqrt{t} + \sqrt{1+t} - 4 \). It combines square roots and constants. We need to understand if and where the function equals zero within the interval \((0, \infty)\).

Step 2: Identify the behavior of the function

First, we observe the behavior at obvious points: At \(t = 0\), \(g(0) = \sqrt{0} + \sqrt{1} - 4 = 1 - 4 = -3\), which is negative. As \(t \to \infty\), \(g(t) = \sqrt{t} + \sqrt{1+t}\), both terms grow without bound making \(g(t)\) positive and increasingly large. Hence, \(g(t)\) goes from negative at \(t=0\) to positive as \(t\) increases.

Step 3: Confirm the behavior change

Since \(g(t)\) is continuous and changes sign from negative to positive as \(t\) increases, by the Intermediate Value Theorem, there must be at least one zero in the interval \((0, \infty)\).

Step 4: Verify if the zero is unique

To determine if this zero is unique, examine the derivative \(g'(t)\):\[ g'(t) = \frac{1}{2\sqrt{t}} + \frac{1}{2\sqrt{1+t}} \].This derivative is always positive for \(t > 0\), indicating \(g(t)\) is strictly increasing for \(t > 0\). Thus, the function can cross the x-axis at most once, confirming the zero is unique.

Key Concepts

Function Behavior Analysis

The first step when dealing with any function is to understand its overall behavior, which allows us to predict how it will act over a certain interval. Let's look at the function given:

• For the function \( g(t) = \sqrt{t} + \sqrt{1+t} - 4 \), we can analyze its behavior through its structure. The components \( \sqrt{t} \) and \( \sqrt{1+t} \) suggest that the function involves growth (as both parts increase with increasing \( t \)).
• Initially, analyze the function at a simple, known point like \( t = 0 \). At \( t = 0 \), calculate the value of the function: \( g(0) = \sqrt{0} + \sqrt{1} - 4 = 1 - 4 = -3 \). This gives us an important point since the function is less than zero here.
• As \( t \) continues to grow towards infinity, \( g(t) = \sqrt{t} + \sqrt{1+t} \) becomes increasingly large since both terms add infinitively, making \( g(t) \) positive. This transition from negative to positive values suggests there's a zero.

This initial analysis is critical as it sets the stage for connecting with the Intermediate Value Theorem.

Continuity and Sign Change

Continuity in functions is essential when trying to apply the Intermediate Value Theorem. The function \( g(t) = \sqrt{t} + \sqrt{1+t} - 4 \) is continuous over the interval \((0, \infty)\) because it is composed of continuous operations: square roots and addition.

To pinpoint the presence of a zero, observe when the function changes signs.

• At \( t = 0 \), the function value \( g(0) = -3 \), shows that it starts negative.
• For large values of \( t \), \( g(t) \) becomes very large and positive.
• Since the function is continuous and we have a sign change from negative to positive as \( t \) increases, the Intermediate Value Theorem guarantees a zero exists somewhere in \((0, \infty)\).

This zero crossing is why the Intermediate Value Theorem is incredibly powerful when dealing with polynomial, radical, or any continuous functions. It's like a bridge that tells us there is at least one place where the function will touch the x-axis.

Derivative Analysis

Once you've established the existence of a zero, it's critical to analyze whether this zero is unique. That's where derivatives come into play. By studying the derivative of a function, we can understand its rate of change. For \( g(t) \), the derivative is given by:

\[ g'(t) = \frac{1}{2\sqrt{t}} + \frac{1}{2\sqrt{1+t}} \]

• This derivative is positive for \( t > 0 \). Why? Because both components \( \frac{1}{2\sqrt{t}} \) and \( \frac{1}{2\sqrt{1+t}} \) are positive wherever their square roots are defined, i.e., for all positive \( t \).
• With \( g'(t) > 0 \), the function \( g(t) \) is strictly increasing on the interval \((0, \infty)\).
• When a function is strictly increasing, it can cross the x-axis at most once within any continuous interval. That's because it doesn't 'turn back' to recross it.

So, derivative analysis confirms not just that a zero exists, but also that there's exactly one zero in the specified interval. This mathematical approach ensures that we understand both the existence and uniqueness of solutions for functions.