Problem 47 - Calculus with Concepts in Calculus

Question

For which functions \(f\) is there a function \(g\) such that $$ f=\sqrt{1+g} $$

Step-by-Step Solution

Answer

Any real-valued function \(f(x)\) has a corresponding \(g(x)\) such that \(f=\sqrt{1+g}\).

Step 1: Recognizing the Function Relationship

The given relationship is \( f(x) = \sqrt{1 + g(x)} \). We need to determine the condition under which a function \( g(x) \) can exist for a given \( f(x) \).

Step 2: Understand the Domain Restriction

Since the expression inside the square root must be non-negative, \( 1 + g(x) \geq 0 \). Thus, \( g(x) \geq -1 \) for every value \( x \). This means \( f(x)^2 = 1 + g(x) \geq 0 \).

Step 3: Solving for g(x)

Rearrange the equation \( f(x) = \sqrt{1 + g(x)} \) to express \( g(x) \) in terms of \( f(x) \). By squaring both sides, we obtain \( f(x)^2 = 1 + g(x) \). Thus, \( g(x) = f(x)^2 - 1 \).

Step 4: Determining Validity Conditions on f

Since \( g(x) \geq -1 \), substitute \( g(x) = f(x)^2 - 1 \) into the inequality: \( f(x)^2 - 1 \geq -1 \). Simplifying gives \( f(x)^2 \geq 0 \), which is always true for any real-valued \( f(x) \). Therefore, any \( f(x) \) can potentially have a corresponding \( g(x) \).

Key Concepts

Domain Restriction

The domain of a function refers to all the input values for which the function is defined. When dealing with square root functions, one must be cautious of the domain restriction because the expression under the square root must be non-negative. This is because you cannot take the square root of a negative number in the realm of real numbers.

In the given problem, the square root function is given as \( f(x) = \sqrt{1 + g(x)} \). To ensure the square root is defined for all \( x \), the expression inside the square root, \( 1+g(x) \), must be greater than or equal to zero.

• Therefore, we conclude that \( g(x) \geq -1 \).

This helps in determining the possible values for \( g(x) \), ensuring that the function \( f(x) \) remains valid in terms of its domain.

Function Relationship

To understand how functions \( f(x) \) and \( g(x) \) are related, examine the equation \( f(x) = \sqrt{1 + g(x)} \). This equation establishes a direct relationship between \( f(x) \) and \( g(x) \).

When you have such a relationship, it often helps to express one function in terms of the other. In our case, this is done by rearranging the terms of the function. By squaring both sides in the step-by-step solution, we derive:

• \( f(x)^2 = 1 + g(x) \)
• Therefore, \( g(x) = f(x)^2 - 1 \)

This tells us that any value of \( f(x) \) directly determines the value of \( g(x) \), which helps in understanding how to construct \( g(x) \) for a given \( f(x) \).

Inequality Condition

Inequality conditions in functions ensure the validity of certain expressions over a range of input values. For this problem, once we rearrange the relation to \( f(x)^2 = 1 + g(x) \), we insert this into the inequality \( g(x) \geq -1 \).

Thus, we get:

• \( f(x)^2 - 1 \geq -1 \)
• Simplifying this inequality gives \( f(x)^2 \geq 0 \)

This condition, \( f(x)^2 \geq 0 \), is always satisfied by real numbers as the square of any real number is non-negative. This reinforces that \( any \) real function \( f(x) \) can potentially have a corresponding \( g(x) \) satisfying the inequality, validating the existence of the function \( g \).

Square Root Function

The square root function involves finding a number that, when multiplied by itself, gives the original number. For the square root expression \( f(x) = \sqrt{1+g(x)} \), it is crucial that the argument (\( 1 + g(x) \)) doesn't turn negative.

The restrictions like \( 1 + g(x) \geq 0 \) ensure the square root function remains within the domain of real numbers. It simplifies mathematical manipulations such as squaring both sides to explore and solve for another function, allowing us to write:

• Initial expression: \( f(x) = \sqrt{1 + g(x)} \)
• Squared form: \( f(x)^2 = 1 + g(x) \)
• Express \( g(x) \) as: \( g(x) = f(x)^2 - 1 \)

This exploration shows how critical understanding square root functions and their constraints is for accurately constructing and analyzing relationships between functions.