Problem 42 - Calculus: Early Transcendentals

Question

Functions with roots Determine the interval(s) on which the following functions are continuous. At which finite endpoints of the intervals of continuity is \(f\) continuous from the left or continuous from the right? $$f(x)=\sqrt{25-x^{2}}$$

Step-by-Step Solution

Answer

Answer: The function f(x) = √(25 - x^2) is continuous on the closed interval [-5, 5].

Step 1: Identify the domain of f(x)

In order for f(x) to be defined, the expression inside the square root must be greater than or equal to zero. So, we need to solve the inequality: $$25 - x^2 \ge 0$$

Step 2: Solving the inequality

To solve this inequality, we can factor the expression inside the square root as follows: $$(5 - x)(5 + x) \ge 0$$ Now we can apply the Zero-Product Property and identify the critical points: $$x = -5, 5$$

Step 3: Determine the intervals of continuity

We know that f(x) is continuous where it is defined. Since the critical points are x = -5 and x = 5, we get the following intervals: $$[-5, 5]$$

Step 4: Determine continuity at endpoints

Now, we need to determine if the function is continuous from the left or right at these finite endpoints: At x = -5, f(x) is continuous from the right, as \(f(x)\) is defined and continuous for \(x \ge -5\). At x = 5, f(x) is continuous from the left, as \(f(x)\) is defined and continuous for \(x \le 5\). To summarize, the function f(x) = \(\sqrt{25 - x^2}\) is continuous on the closed interval [-5, 5]. It is continuous from the right at x = -5 and continuous from the left at x = 5.

Key Concepts

Domain of a Function

When we talk about the domain of a function, we are referring to all possible input values, usually denoted as "x", for which the function is defined. For example, in the function \(f(x) = \sqrt{25 - x^2}\), the function only exists for values of \(x\) where the expression under the square root is non-negative. This is because the square root of a negative number is not a real number.

To find this domain, we solve the inequality \(25 - x^2 \ge 0\). Solving this expression intends to find the set of x-values that make \(25 - x^2\) zero or positive. Mathematically, this means finding intervals where the function is real and therefore defined.

• The roots of the equation \(25 - x^2 = 0\) are \(x = 5\) and \(x = -5\). These roots split the real number line into different segments.
• Testing values from each segment determines the sign of the inequality, allowing us to know where the function is positive, and hence, real and defined.

For the function \(f(x) = \sqrt{25 - x^2}\), its domain is \([-5, 5]\), meaning it's defined for all x-values from -5 to 5 inclusive.

Intervals of Continuity

Intervals of continuity are the stretches of the function's domain where the function does not "break" or "jump" in value. In other words, a function is continuous on an interval when there are no sudden changes in output values as the input transitions through that interval.

For \(f(x) = \sqrt{25-x^{2}}\), we established that the domain is \([-5, 5]\). Since square root functions are continuous wherever they are defined, \(f(x)\) is continuous through the entire interval from -5 to 5.

• When x is within this interval, the function smoothly transitions from one value to another, which means it does not have undefined points, jumps, or breaks.
• The continuity also considers endpoint behaviors, meaning we look at how the function behaves as it approaches the edge of this interval.

Specifically, as \(x\) approaches -5 from the right and as \(x\) approaches 5 from the left, the function remains defined and connected, ensuring it is right-continuous at -5 and left-continuous at 5.

Zero-Product Property

The Zero-Product Property is a fundamental concept used in solving equations of the form \(ab=0\), where you conclude that either \(a=0\) or \(b=0\) and sometimes both. In our analysis of \(f(x) = \sqrt{25-x^2}\), this principle is instrumental in identifying where the inside of the square root is zero.

We factored \(25-x^2\) into \((5-x)(5+x)\). According to the Zero-Product Property, this product equals zero when either factor equals zero.

• Setting \(5-x = 0\) results in \(x = 5\).
• Setting \(5+x = 0\) results in \(x = -5\).

These values, \(x = -5\) and \(x = 5\), are crucial as they mark the boundaries of the intervals where \(f(x)\) is continuous. They are the roots of the original inequality \(25 - x^2 \ge 0\), thus indicating the edges of the domain \([-5, 5]\). This property ensures that the analyzed function transitions through these critical points in an orderly, defined manner.