Question

Determine the largest interval over which the given function is continuous. $$ f(x)=\frac{1}{\sqrt{25-x^{2}}} $$

Short answer

The largest interval over which the function is continuous is \( (-5, 5) \).

Step-by-step solution

Identify the expression under the square root

The function given is \( f(x) = \frac{1}{\sqrt{25-x^{2}}} \). Notice that the expression \( \sqrt{25-x^{2}} \) is under the square root, and a square root is only defined for non-negative numbers.

Set condition for square root

For the square root \( \sqrt{25-x^{2}} \) to be real, the expression inside must be greater than zero: \( 25 - x^{2} > 0 \). This is because it is in the denominator and cannot be zero.

Solve the inequality

Solve the inequality \( 25 - x^{2} > 0 \) to find the values of \( x \). First, rearrange the inequality: \( 25 > x^{2} \). Taking the square root of both sides, we get \( -5 < x < 5 \). This gives us the interval where the function is defined.

Interpret the result

The function \( f(x) \) is continuous wherever it is defined. Since the square root expression must be positive and never zero to keep the function defined, it is continuous over the interval \( -5 < x < 5 \). The endpoints \( x = -5 \) and \( x = 5 \) are not included in the interval because the function approaches infinity and is not defined there.

Key concepts

Interval of Continuity

When exploring the concept of continuity, it is important to consider where a function remains unbroken or "continuous" as you move across the x-axis. For the function in question, \( f(x) = \frac{1}{\sqrt{25-x^{2}}} \), continuity means that the function should be defined and smooth for all values within a certain range, known as the interval of continuity.
In this exercise, the continuity of the function is determined by the conditions set by the square root and the denominator. Since square roots are only defined for non-negative numbers and the expression \( 25-x^{2} \) lies in the denominator, it must be greater than zero for the function to be valid. This leads us to solving the inequality \( 25 - x^{2} > 0 \), which ultimately gives us the interval \( -5 < x < 5 \).
Thus, the function is continuous over the interval \( -5 < x < 5 \), and this interval represents all values of \( x \) where the function does not experience any jumps, holes, or vertical asymptotes.

Inequalities in Calculus

Inequalities are pivotal in calculus as they help establish the boundaries within which functions behave in certain ways, like being continuous. In this problem, we encountered the inequality expression \( 25 - x^{2} > 0 \). Solving this inequality helps us find the range of x-values where the square root expression remains positive, ensuring the function is well-defined.
Here’s a simple breakdown of solving this inequality:

• Rewrite the inequality: \( 25 > x^{2} \).
• Take the square root to find the solutions: \( 5 > |x| \).
• Break it down to: \( -5 < x < 5 \).

This solution shows the interval where the function maintains its continuity and where any input infers a real number output. Solving inequalities like these help define exactly where the function’s behavior is predictable and unbroken by changing input values.

Squares and Square Roots

Understanding squares and square roots is critical when dealing with functions involving these expressions. In the given function \( f(x) = \frac{1}{\sqrt{25-x^{2}}} \), the square root is a crucial component that determines where the function is defined. Square roots are only defined for non-negative numbers, hence requiring attention to the expression inside them.
Let's break down a few key points about square roots:

• The expression under a square root must be \( \geq 0 \) for real number outcomes.
• In the denominator, a expression under a square root must be \( > 0 \) to prevent division by zero.
• Taking square roots can help simplify inequalities, enabling us to find valid intervals.

The square root \( \sqrt{25-x^{2}} \) implies that \( 25-x^{2} > 0 \). Solving gives us the interval \( -5 < x < 5 \). Understanding the behavior of both squares and square roots is fundamental to correctly evaluate and work with equations and inequalities in calculus.