Problem 16 - Calculus for Biology and Medicine

Question

In Problems 15-24, find the values of \(x \in\) R for which the given functions are both defined and continuous. $$ f(x)=\sqrt{x^{2}-1} $$

Step-by-Step Solution

Answer

The function is defined and continuous for \( x \in (-\infty, -1] \cup [1, \infty) \).

Step 1: Understand the Function

The function given is \( f(x) = \sqrt{x^2 - 1} \). It is a square root function, which means that the expression inside the square root, \( x^2 - 1 \), must be greater than or equal to zero for \( f(x) \) to be defined.

Step 2: Solve the Inequality

Set the inequality for the expression inside the square root: \[ x^2 - 1 \geq 0 \]. This inequality can be rearranged to: \[ x^2 \geq 1 \].

Step 3: Solve for x

To solve \( x^2 \geq 1 \), consider the two possible cases:1. \( x \geq 1 \)2. \( x \leq -1 \)This is because, for any positive number on the left side of the inequality, the absolute value will have two possible solutions in linear inequalities.

Step 4: Verify Domain and Continuity

For the function \( f(x) \) to be both defined and continuous, the value under the square root must be non-negative and continuous on the intervals \( x \in (-\infty, -1] \cup [1, \infty) \). Both conditions hold for these intervals because square root functions are continuous where they are defined.

Key Concepts

Square Root Function

A square root function typically involves the square root of an algebraic expression. In our case, the function is expressed as \( f(x) = \sqrt{x^2 - 1} \).
To ensure that this function is defined, the expression inside the square root, \( x^2 - 1 \), must be non-negative, meaning it has to be either zero or positive. This condition is necessary because the square root of a negative number is not defined within the set of real numbers.

• When analyzing square root functions, first identify the expression under the square root.
• Ensure that this expression is greater than or equal to zero so that \( f(x) \) is properly defined.

Square root functions are naturally continuous wherever they are defined, given that no jumps or breaks can occur in their graph.

Inequality Solutions

To find where \( f(x) = \sqrt{x^2 - 1} \) is defined, we must solve the inequality \( x^2 - 1 \geq 0 \). Solving this requires us to determine where \( x^2 \) is at least 1.
This is solved by considering two cases, because the inequality has two roots, \( x^2 = 1 \), or \( x = \pm 1 \). These points divide the number line into intervals which we can individually assess to determine where the inequality holds true:

• For \( x \geq 1 \), the values of \( x^2 \) are naturally greater than or equal to 1.
• For \( x \leq -1 \), the values of \( x^2 \) remain greater than or equal to 1 due to the squares of negative numbers returning positive results.

Thus, the inequality is satisfied in the intervals \((-\infty, -1] \cup [1, \infty)\), which represents the range of \( x \) for which the function is defined and properly handles real inputs.

Function Domain

The domain of a function consists of all the input values (in this case, \( x \)) for which the function yields real, valid outputs.
For the square root function \( f(x) = \sqrt{x^2 - 1} \), the domain is determined by ensuring the expression under the square root is non-negative.

• The inequality \( x^2 - 1 \geq 0 \) tells us the valid domain for the function.
• This results in the domain \( x \in (-\infty, -1] \cup [1, \infty) \).

Within this domain, the function is also continuous, as there are no discontinuities or breaks. Therefore, in these intervals, every input leads to a valid, real output for function \( f(x) \). Remember, evaluating the domain effectively helps us understand where the function behaves well and remains defined.