Problem 50 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 49-54, determine the largest interval over which the given function is continuous. $$ f(x)=\frac{1}{\sqrt{25-x^{2}}} $$

Step-by-Step Solution

Answer

The function is continuous on the interval \((-5, 5)\).

Step 1: Identify Conditions for Continuity

A function is continuous where it is defined. The function \( f(x) = \frac{1}{\sqrt{25-x^2}} \) is continuous where the denominator is non-zero and defined. Consequently, we need \( \sqrt{25-x^2} > 0 \).

Step 2: Solve for the Interval of x

The inequality \( \sqrt{25-x^2} > 0 \) implies that \( 25-x^2 > 0 \). This can be rewritten as \( 25 > x^2 \). Solving for \( x \), this gives \( -5 < x < 5 \).

Step 3: Determine the Largest Interval

The solution to \( -5 < x < 5 \) describes the largest interval over which the function is continuous, as it ensures the expression under the square root is positive and the denominator is not zero.

Key Concepts

Continuity Conditions

Continuity is a fundamental concept in calculus that describes how smoothly a function behaves. At a basic level, a function is said to be continuous at a point if there are no sudden jumps or breaks in the graph at that point. For a function like \( f(x) = \frac{1}{\sqrt{25-x^2}} \), determining where it's continuous involves ensuring the denominator is never zero or undefined.

• **Defined**: The function must exist at every point in the interval.
• **Non-Zero Denominator**: The function is only continuous where the denominator, \( \sqrt{25-x^2} \), is greater than zero. Not equal to zero as this would make the function undefined.
• **Without Breaks**: The function should not have sudden changes in its value.

To ensure these conditions, we solve for when \( \sqrt{25-x^2} > 0 \). This leads directly to determining the set of \( x \) values that keep the square root positive, thus ensuring continuity.

Inequalities

Inequalities are mathematical expressions that allow us to state that one quantity is larger or smaller than another. In our context, we used the inequality \( \sqrt{25-x^2} > 0 \) to explore where the function \( f(x) = \frac{1}{\sqrt{25-x^2}} \) remains continuous.
### Solving InequalitiesTo solve \( \sqrt{25-x^2} > 0 \):1. **Square Both Sides**: This removes the square root. Now the inequality is \( 25-x^2 > 0 \).2. **Rearrange Terms**: Bringing \( x^2 \) to the other side gives \( x^2 < 25 \).3. **Solve for x**: * Find the positive and negative square roots of 25: \(-5 < x < 5\).Through solving this inequality, we determine the values of \( x \) that keep the function defined and continuous.

Interval Notation

Interval notation is a concise way to describe a set of numbers between given endpoints. It is especially useful when we want to specify the domain over which a function, such as \( f(x) = \frac{1}{\sqrt{25-x^2}} \), is continuous.
### Using Interval NotationWhen we find that \( -5 < x < 5 \):

• **Open Interval**: The interval is written as \((-5, 5)\). This means \( x \) can take any value between \( -5 \) and \( 5 \), excluding \( -5 \) and \( 5 \) themselves.
• **Parentheses**: The use of parentheses \(()\) signifies that the endpoints are not included. This aligns with the inequality as the function is undefined at exactly \( x = -5 \) and \( x = 5 \).

This notation tells us quickly and clearly the range of \( x \) values making the function continuous. Understanding interval notation is key to expressing domains and ranges efficiently in mathematics.