Problem 3 - The Calculus with Analytic Geometry

Question

Determine whether the function is continuous or discontinuous on each of the indicated intervals. $$ g(x)=\sqrt{x^{2}-9} ;(-\infty,-3),(-\infty,-3],(3,+\infty),[3,+\infty),(-3,3) $$

Step-by-Step Solution

Answer

Continuous in \((-\rightarrowfty, -3)\), \((-\rightarrowfty, -3]\), \((3, +\rightarrowfty)\), \([3, +\rightarrowfty)\). Discontinuous in \((-3, 3)\).

Step 1: Understand the function

The given function is \( g(x) = \sqrt{x^2 - 9} \). The square root function is defined for non-negative arguments, so \( x^2 - 9 \geq 0 \).

Step 2: Determine the domain

The inequality \( x^2 - 9 \geq 0 \) holds true when \( x \leq -3 \) or \( x \geq 3 \). Therefore, the domain of the function is \((-\infty, -3] \cup [3, +\infty)\).

Step 3: Interval \((-\rightarrowfty, -3)\)

In the interval \((-\rightarrowfty, -3)\), the function is continuous because it is part of the domain where \( x \leq -3 \).

Step 4: Interval \((-\rightarrowfty, -3]\)

In the interval \((-\rightarrowfty, -3]\), the function is also continuous because it includes the endpoint \( -3 \), which is within the domain of the function.

Step 5: Interval \((3, +\rightarrowfty)\)

In the interval \((3, +\rightarrowfty)\), the function is continuous because it is part of the domain where \( x \geq 3 \).

Step 6: Interval \([3, +\rightarrowfty)\)

In the interval \([3, +\rightarrowfty)\), the function remains continuous because it includes the endpoint \( 3 \), which is within the domain.

Step 7: Interval \((-3, 3)\)

In the interval \((-3, 3)\), the function is discontinuous because the values between \(-3\) and \(3\) are not within the domain of the function (since \( x^2 - 9 < 0 \)).

Key Concepts

Function Domain

The domain of a function is the set of all possible input values (x-values) that make the function work without causing any mathematical issues like division by zero or taking the square root of a negative number.
For the function \( g(x) = \sqrt{x^2 - 9} \), we need to find where the expression inside the square root is non-negative.
This means solving the inequality \( x^2 - 9 \geq 0 \).
To solve, factorize the inequality: \((x - 3)(x + 3) \geq 0\).
You get two critical points: \( x = -3 \) and \( x = 3 \).
This inequality holds true when \( x \leq -3 \) or \( x \geq 3 \).
Therefore, the domain of \( g(x) \) is \( (-\infty, -3] \cup [3, +\infty) \).
These are the only x-values where the function can exist without running into mathematical errors. Always verify the domain of a function before evaluating it at specific points. This foundational step ensures you're working within permissible boundaries.

Discontinuous Intervals

A function is discontinuous where it is not defined or where it has abrupt changes. Let's consider the function \( g(x) = \sqrt{x^2 - 9} \).
The domain of this function shows us where it is defined, but we also use it to identify discontinuities.
This function is discontinuous in the interval \((-3, 3)\) because any value of x within this range causes \( x^2 - 9 \) to become negative.
Therefore, for \( g(x) \) to be continuous, it should not have breaks, jumps, or holes in these intervals. Make sure to always check the inside expression of square roots and denominators to determine these intervals. For \( g(x) \), any value between -3 and 3 makes the argument of the square root negative, hence the function cannot exist, causing discontinuity.

Continuous Intervals

Continuity in a function means there are no breaks, jumps, or holes in the graph. Let's examine where \( g(x) \) is continuous based on the domain.
The intervals \((-\infty, -3] \) and \( [3, +\infty) \) are within the domain of the function.
Therefore, \( g(x) \) is continuous in these intervals as it experiences no interruptions or undefined points. When working within these intervals, the function is smoothly connected without any abrupt changes. Checking for continuous intervals ensures clarity about where a function behaves predictably, and helps avoid missteps in more advanced calculations. Always double-check continuous intervals by ensuring the function is well-defined and behaves regularly without any sudden breaks or jumps.