Problem 46 - Calculus with Concepts in Calculus

Question

Let \(f(x)=\sqrt{x^{2}+1}-1\) and \(g(x)=\frac{x^{2}}{1+\sqrt{x^{2}+1}}\) a. Find the domains of \(f\) and \(g\). b. Show that \(f=g\).

Step-by-Step Solution

Answer

Domains of \( f(x) \) and \( g(x) \) are \((-\infty, \infty)\), and \( f(x) = g(x) \) for all \( x \).

Step 1: Determine the Domain of f(x)

To find the domain of \( f(x) = \sqrt{x^2 + 1} - 1 \), we need to ensure that the expression inside the square root is non-negative. Since \( x^2 \) is always non-negative and \( x^2 + 1 \) is always positive for all real numbers \( x \), the domain of \( f(x) \) is all real numbers, which can be written as \( (-\infty, \infty) \).

Step 2: Determine the Domain of g(x)

For \( g(x) = \frac{x^2}{1 + \sqrt{x^2 + 1}} \), the denominator must not be zero. The expression \( 1 + \sqrt{x^2 + 1} \) is always positive because both terms are positive: \( \sqrt{x^2 + 1} \geq 1 \). Therefore, the domain of \( g(x) \) is also all real numbers \( (-\infty, \infty) \).

Step 3: Simplify f(x)

Let's rewrite \( f(x) = \sqrt{x^2 + 1} - 1 \). Multiply and divide by the conjugate \( \sqrt{x^2 + 1} + 1 \) to simplify: \[f(x) = (\sqrt{x^2 + 1} - 1)\times\frac{\sqrt{x^2 + 1} + 1}{\sqrt{x^2 + 1} + 1} = \frac{(\sqrt{x^2 + 1})^2 - 1^2}{\sqrt{x^2 + 1} + 1} = \frac{x^2}{1 + \sqrt{x^2 + 1}}.\]This is the expression for \( g(x) \).

Step 4: Confirm f(x) = g(x)

Since \( f(x) \) simplifies to \( \frac{x^2}{1 + \sqrt{x^2 + 1}} \), it matches the expression for \( g(x) \). This confirms that \( f(x) = g(x) \) for all \( x \) in the domain, which is all real numbers \((-\infty, \infty)\).

Key Concepts

Domain of a Function

The domain of a function is the set of all possible inputs (or 'x' values) that a function can accept without breaking any mathematical rules. It's like finding out all the x-values you can plug into a function and still get an answer.
To figure out the domain, we need to consider things like:

• Square roots, which can only have non-negative numbers inside.
• Denominators that cannot be zero in fractions.

For the function \( f(x) = \sqrt{x^2 + 1} - 1 \), the expression inside the square root, \( x^2 + 1 \), is always positive. This is because squaring any real number results in a non-negative value, and adding 1 makes it always positive no matter what x is.
Similarly, for \( g(x) = \frac{x^2}{1 + \sqrt{x^2 + 1}} \), the denominator \( 1 + \sqrt{x^2 + 1} \) is never zero, as \( \sqrt{x^2 + 1} \) is always at least 1.
This means both functions have the domain of all real numbers: \( (-\infty, \infty) \).

Simplifying Expressions

Simplifying expressions involves rewriting them in a form that is easier to work with. Sometimes this means manipulating the expression until it resembles something more familiar or terser.
For example, consider \( f(x) = \sqrt{x^2 + 1} - 1 \). A strategy for simplifying this is to multiply by the conjugate, which is a technique particularly useful when dealing with square roots around fractions."

• The conjugate of \( \sqrt{x^2 + 1} - 1 \) is \( \sqrt{x^2 + 1} + 1 \).
• Multiplying and dividing by \( \sqrt{x^2 + 1} + 1 \) simplifies \( f(x) \) to a function form identical to \( g(x) \): \( \frac{x^2}{1 + \sqrt{x^2 + 1}} \).

This simplification shows how factors cancel out or result in simpler numerators and denominators, helping to verify certain properties or equate functions more facilely.

Equality of Functions

Understanding when two functions are equal involves ensuring that for every x-value in their domain, their output is identical.
This means checking their simplified forms are exactly alike all the way through their domain.
In the exercise, we worked out that \( f(x) \) simplifies to \( \frac{x^2}{1 + \sqrt{x^2 + 1}} \), which is exactly \( g(x) \).
Since both \( f(x) \) and \( g(x) \) are defined for all real numbers and produce the same output for any x-value, they are equal on that domain. That means \( f(x) = g(x) \), showcasing function equivalence when one has been simplified properly.