Problem 21 - Thomas Calculus

Question

Show that the functions in Exercises \(19-26\) have exactly one zero in the given interval. \begin{equation}g(t)=\sqrt{t}+\sqrt{1+t}-4, \quad(0, \infty)\end{equation}

Step-by-Step Solution

Answer

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

Step 1: Understand the function

We are given the function \( g(t) = \sqrt{t} + \sqrt{1+t} - 4 \) and need to show that it has exactly one zero in the interval \((0, \infty)\). This means that there is exactly one solution \(t \in (0, \infty)\) for which \(g(t) = 0\).

Step 2: Analyze the endpoints

At \(t=0\), \(g(0) = \sqrt{0} + \sqrt{1+0} - 4 = 1 - 4 = -3\). As \(t \to \infty\), both \(\sqrt{t}\) and \(\sqrt{1+t}\) increase indefinitely, so \(g(t) \to \infty\). Thus, \(g(t)\) changes from negative to positive as \(t\) goes from 0 to \(\infty\).

Step 3: Verify continuity of the function

The function \( g(t) = \sqrt{t} + \sqrt{1+t} - 4 \) is continuous for all \( t \geq 0 \) because it is the sum of continuous functions \( \sqrt{t} \) and \( \sqrt{1+t} \), subtracting a constant does not affect continuity.

Step 4: Apply Intermediate Value Theorem

Since \(g(0) = -3 < 0\) and \(g(t) \to \infty\) as \(t \to \infty\) and \(g(t)\) is continuous, by the Intermediate Value Theorem, there exists at least one \(c \in (0, \infty)\) such that \(g(c) = 0\).

Step 5: Prove the uniqueness of the solution

To show uniqueness, compute \(g'(t) = \frac{1}{2\sqrt{t}} + \frac{1}{2\sqrt{1+t}}\). This derivative \(g'(t) > 0\) for all \(t > 0\), indicating that \(g(t)\) is strictly increasing in \( (0, \infty) \). Thus, \(g(t)\) can have at most one zero in the interval \((0, \infty)\).

Step 6: Conclusion

Since \(g(t)\) is continuous over \((0, \infty)\), changes from negative to positive, and is strictly increasing, it has exactly one zero in \((0, \infty)\).

Key Concepts

Continuous Functions

A continuous function is a type of function where small changes in the input lead to small changes in the output. This means there are no sudden jumps or holes in the graph of the function. For a function to be continuous at a point, it must be that as you approach that point from either side, the function approaches the same value.
In our exercise, the function given is \( g(t) = \sqrt{t} + \sqrt{1+t} - 4 \). This function is continuous over \( t \geq 0 \) because both \( \sqrt{t} \) and \( \sqrt{1+t} \) are continuous functions. Adding or subtracting constants and other continuous functions maintains this continuity.

• The square root function, \( \sqrt{t} \), is continuous for all non-negative \( t \).
• The function \( \sqrt{1+t} \) is continuous for all real numbers\( t \).
• Subtraction of constants (in this case, 4) does not affect continuity.

The Intermediate Value Theorem relies on the concept of continuity. It states that for any value between two endpoints of a continuous function, there is at least one point within the interval where the function will take that value.

Strictly Increasing Functions

A strictly increasing function is a function where, as the input increases, the output also increases. This means that for any two numbers \( a \) and \( b \) where \( a < b \), it follows that \( f(a) < f(b) \). This property ensures that the function does not repeat any values, which is crucial for proving uniqueness of solutions.
In the solution of our exercise, we calculate that the derivative of the function \( g(t) \) is \( g'(t) = \frac{1}{2\sqrt{t}} + \frac{1}{2\sqrt{1+t}} \). This derivative is positive for all \( t > 0 \) because both terms \( \frac{1}{2\sqrt{t}} \) and \( \frac{1}{2\sqrt{1+t}} \) are positive when \( t > 0 \).

• This ensures that \( g(t) \) is a strictly increasing function in the interval \( (0, \infty) \).
• Thus, the function cannot have more than one zero because a strictly increasing continuous function can cross the horizontal axis at most once.

By showing the function is strictly increasing, we conclude it can only intersect the x-axis one time in the interval \( (0, \infty) \), guaranteeing the uniqueness of the zero.

Real-Valued Functions

Real-valued functions are functions where the output is a real number for every input in its domain. This concept is essential because it ensures the Intermediate Value Theorem applies, providing us a way to find zeros of the function.
For our function \( g(t) = \sqrt{t} + \sqrt{1+t} - 4 \), both \( \sqrt{t} \) and \( \sqrt{1+t} \) return real numbers for non-negative inputs. The difference here, \( -4 \), is also a real number. Thus, \( g(t) \) is real-valued for all \( t \geq 0 \).

• Real-valued functions are necessary when applying the Intermediate Value Theorem, which requires the function to take on all values between \( g(0) \) and its behavior as \( t \to \infty \).
• A real-valued function allows us to evaluate its derivative and confirm properties such as being strictly increasing.

By understanding that \( g(t) \) is real-valued, we confirm it's suitable for applying theoretical concepts like the Intermediate Value Theorem and ensure precise mathematical evaluation through its derivative.