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Chapter 1: Problem 113

For the following exercises, find the domain of each function using interval notation. $$\frac{2 x+1}{\sqrt{5-x}}$$

Short Answer

Expert verified
The domain is \((-\infty, 5]\).

Step by step solution

01

Identify the function

The function given is \( f(x) = \frac{2x+1}{\sqrt{5-x}} \). We are tasked with finding its domain.
02

Understand the constraints for the square root

A square root function, \( \sqrt{a} \), exists for non-negative values only. Therefore, \( 5-x \geq 0 \) must be true to ensure the square root exists.
03

Solve the inequality

Solve the inequality \( 5 - x \geq 0 \). By adding \( x \) to both sides, we get \( 5 \geq x \). Thus, \( x \) must be less than or equal to 5.
04

Consider all values included

From the inequality, all \( x \) values that satisfy \( x \leq 5 \) are included. The expression in the numerator, \( 2x+1 \), does not impose additional restrictions, as it is defined for all real numbers.
05

Write the domain in interval notation

Since \( x \) must be less than or equal to 5 and there are no further restrictions on negative numbers (as long as the square root is defined), we write the domain in interval notation as \((-\infty, 5]\).

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Key Concepts

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

Interval Notation
Interval notation is a system used to describe a set of numbers, which is particularly useful to specify the domain of a function. This helps articulate the specific range of "x-values" that a function can accept while being valid.
When writing interval notation:
  • Parentheses \((a, b)\) signify that the endpoints \(a\) and \(b\) are not included (open interval).
  • Brackets \([a, b]\) signify that the endpoints \(a\) and \(b\) are included (closed interval).
  • You can use these symbols in various combinations to describe whether the interval includes its endpoints, like \((a, b]\) or \([a, b)\).
  • The symbol \(-\infty\) is always paired with a parenthesis since infinity is a concept, not a number that can be included.
In the example given, the function's domain is all real x-values below or equal to 5. Thus, the correct interval notation is \(( -\infty, 5] \). This notation tells us every possible real number up to and including 5 is valid.
Inequalities
Inequalities are mathematical expressions involving the symbols \(<\), \(>\), \(\leq\), and \(\geq\). They express the relative size or order of two values.
Usually, they help in determining the limits or constraints for a function, as they show which values make the expression true.
  • The expression \(a < b\) means \(a\) is less than \(b\).
  • The expression \(a \leq b\) means \(a\) is less than or equal to \(b\).
  • They can also be used in solving conditions under which a particular function remains valid, which is crucial for defining domains.
In our function \(f(x) = \frac{2x+1}{\sqrt{5-x}}\), the inequality \(5 - x \geq 0\) ensures the square root is defined only when \(x \leq 5\). Any \(x\) beyond this range would render the square root undefined, making the function invalid.
Square Root Constraints
Square root constraints focus on the principle that roots of real numbers must be non-negative to be defined. This principle is vital in understanding how to restrict the input of functions that include square roots.
The expression under a square root must be non-negative because:
  • The square root of a negative number is not a real number.
  • For a function \(\sqrt{a}\) to be defined, \(a \geq 0\).
  • Ignoring this could result in attempting to calculate an undefined or imaginary number.
In the exercise, the term \(\sqrt{5-x}\) in the denominator implies that \(5-x\) must satisfy the condition \(5-x \geq 0\). Solving this inequality, we find \(x \leq 5\). The entire purpose of square root constraints is to ensure we never attempt to calculate square roots of negative numbers, keeping the function within the domain of real numbers.

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Most popular questions from this chapter

For the following exercises, describe how the formula is a transformation of a toolkit function. Then sketch a graph of the transformation. $$ h(x)=-2|x-4|+3 $$

For the following exercises, use the function values for \(f\) and \(g\) shown in Table 1.23 to evaluate each expression. $$ \begin{array}{|c|c|c|c|c|c|c|c|c|}\hline x & {0} & {1} & {2} & {3} & {4} & {5} & {6} & {7} & {8} & {9} \\ \hline f(x) & {7} & {6} & {5} & {8} & {4} & {0} & {2} & {1} & {9} & {3} \\ \hline g(x) & {9} & {5} & {6} & {2} & {1} & {8} & {7} & {3} & {4} & {0} \\ \hline\end{array} $$ $$ g(f(5)) $$

For the following exercises, use each set of functions to find \(f(g(h(x))) .\) Simplify your answers. Given \(f(x)=\frac{1}{x}\) and \(g(x)=x-3,\) find the following: a. \((f \circ g)(x)\) b. the domain of \((f \circ g)(x)\) in interval notation c. \((g \circ f)(x)\) d. the domain of \((g \circ f)(x)\) e. \(\left(\frac{f}{g}\right) x\)

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For the following exercises, write a formula for the function \(g\) that results when the graph of a given toolkit function is transformed as described. The graph of \(f(x)=|x|\) is reflected over the \(y\) -axis and horizontally compressed by a factor of \(\frac{1}{4}\)
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