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Chapter 1: Problem 24

Give the intervals on which the given function is continuous. $$ g(x)=\sqrt{x^{2}-4} $$

Short Answer

Expert verified
The function is continuous on \((-\infty, -2]\) and \([2, \infty)\).

Step by step solution

01

Understanding the Domain of a Square Root Function

The function in the problem is a square root function, specifically \( g(x) = \sqrt{x^2 - 4} \). A square root function is defined for non-negative values of the expression inside the square root. Thus, \( x^2 - 4 \geq 0 \).
02

Solving the Inequality

To find for which values of \( x \) the inequality \( x^2 - 4 \geq 0 \) holds, consider the equation \( x^2 - 4 = 0 \). Solving this, \( x^2 = 4 \), we find \( x = \pm 2 \). These are the points where the function changes behavior. Check the signs of \( x^2 - 4 \) in the intervals \( (-\infty, -2) \), \( (-2, 2) \), and \( (2, \infty) \).
03

Testing Intervals

For \( x < -2 \), for example, \( x = -3 \), substituting gives \( (-3)^2 - 4 = 9 - 4 = 5 > 0 \). Thus, the function is defined for \( x < -2 \). For \( x \) between \(-2\) and \(2\), any point such as \( x = 0 \) gives \( 0^2 - 4 = -4 < 0 \), meaning the function is not defined. For \( x > 2 \), for instance, \( x = 3 \), results in \( 3^2 - 4 = 9 - 4 = 5 > 0 \). Thus, the function is defined. Hence it is defined in \( (-\infty, -2) \) and \( (2, \infty) \).
04

Conclusion

Since \( g(x) = \sqrt{x^2 - 4} \) is continuous wherever it is defined (given no discontinuities at \( x=2 \) and \( x=-2 \)), the function is continuous on the intervals \( (-\infty, -2] \) and \( [2, \infty) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Domain of a Function
The domain of a function defines all possible input values (x-values) for which the function is defined. When dealing with a square root function like \( g(x) = \sqrt{x^2 - 4} \), it is essential to ensure that the expression inside the square root is not negative.
This is because the square root of a negative number is not defined within the set of real numbers.

To find the domain, solve the inequality \( x^2 - 4 \geq 0 \) to determine which values of \( x \) make the expression non-negative.
So, always make sure to check under the square root to find out where the function exists over the real numbers.
  • The domain of \( g(x) \) is derived by setting \( x^2 - 4 \) to be greater than or equal to 0.
Exploring Inequality
Inequalities are mathematical expressions that use greater than \((>)\), less than \((<)\), greater than or equal to \((\geq)\), and less than or equal to \((\leq)\) symbols to compare values or expressions. In the context of the function \( g(x) = \sqrt{x^2 - 4} \), the inequality to solve is \( x^2 - 4 \geq 0\) to find which values of \( x \) keep the expression non-negative.
Solving the equation \( x^2 - 4 \) involves:
  • Factoring or using a zero-product property to find boundary points \( x = 2 \) and \( x = -2 \).
  • Dividing the number line into intervals based on these critical points.
  • Testing each interval to determine where the inequality holds true, ensuring \( x^2 - 4 \) remains non-negative.
Remember, solving inequalities correctly is key to determining valid sections of the domain.
Inspecting Intervals
Intervals are portions of the number line that describe the valid x-values for a function based on the domain. Once we determine the solutions to our inequality \( x^2 - 4 \geq 0 \), we check the resulting intervals to ensure where our function is defined.
Critical points at \( x = -2 \) and \( x = 2 \) divide the x-values into sections.
  • First interval: \((-\infty, -2)\) 鈥� Testing a point (e.g., \( x = -3 \)) in this interval shows that the function outputs a non-negative value, so the function is defined here.
  • Second interval: \((-2, 2)\) 鈥� Testing here with a point like \( x = 0 \), you get a negative outcome in the expression, so the function is undefined in this interval.
  • Third interval: \((2, \infty)\) 鈥� Points like \( x = 3 \) confirm the expression is non-negative, allowing for the function's definition.
Knowing how to correctly test and analyze these intervals helps us accurately describe where the function behaves properly.
Exploring the Square Root Function
Square root functions are expressions where a variable is inside a square root sign. The general form is \( \sqrt{expression}\). For these functions, the inside expression must be zero or positive else it wouldn't result in a real number output.
Understanding the characteristics of a square root function is crucial since:
  • Square roots involving negative numbers yield complex numbers, which aren't considered here for real-valued functions.
  • The simplest square root function is \( \sqrt{x} \), defined only for \( x \geq 0 \).
  • Variations like \( \sqrt{x^2 - 4} \) show that within this context, the behavior changes based on the expression's non-negativity.
This knowledge helps us determine the function's domain and, by extension, its continuity over real numbers.

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Most popular questions from this chapter

Determine if \(f\) is continuous at the indicated values. If not, explain why. \(f(x)=\left\\{\begin{array}{cl}\frac{x^{2}+5 x+4}{x^{2}+3 x+2} & x \neq-1 \\\ 3 & x=-1\end{array}\right.\) (a) \(x=-1\) (b) \(x=10\)

Give the intervals on which the given function is continuous. $$ f(x)=\sin \left(e^{x}+x^{2}\right) $$

Evaluate the given limits of the piecewise defined functions \(f\). $$ \begin{array}{ll} f(x)=\left\\{\begin{array}{cc} 2 x^{2}+5 x-1 & x<0 \\ \sin x & & x \geq 0 \end{array}\right. \\ \begin{array}{ll} \text { (a) } \lim _{x \rightarrow 0^{-}} f(x) & \text { (c) } \lim _{x \rightarrow 0} f(x) \\ \text { (b) } \lim _{x \rightarrow 0^{+}} f(x) & \text { (d) } f(0) \end{array} \end{array} $$

Challenge your understanding of limits but can be evaluated using the knowledge gained in this section. $$ \lim _{x \rightarrow 0} \frac{\ln (1+x)}{x} $$

Give the intervals on which the given function is continuous. $$ h(t)=\cos t $$
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