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Chapter 9: Problem 29

Specify the domain for each of the functions. $$f(s)=\sqrt{4 s-5}$$

Short Answer

Expert verified
The domain of the function is \( \left[\frac{5}{4}, \infty\right) \).

Step by step solution

01

Understand the Function and Its Requirements

The function given is \( f(s) = \sqrt{4s - 5} \). Since the function involves a square root, the expression inside the square root must be non-negative. This is because the square root of a negative number is not defined in the set of real numbers.
02

Set the Expression Inside the Square Root to be Non-Negative

To ensure that the expression inside the square root is non-negative, set up the inequality: \[ 4s - 5 \geq 0 \]This inequality ensures that 4s - 5 is zero or positive.
03

Solve the Inequality for s

Solve the inequality from Step 2:\[4s - 5 \geq 0 \4s \geq 5 \s \geq \frac{5}{4}\]The solution \( s \geq \frac{5}{4} \) are the values for which the function is defined.
04

Write the Domain in Interval Notation

The domain of the function is all real numbers \( s \geq \frac{5}{4} \). In interval notation, this is expressed as:\[ \left[\frac{5}{4}, \infty\right) \]This represents all real numbers starting from \( \frac{5}{4} \) to infinity.

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Key Concepts

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

Inequality solving
Solving inequalities is a fundamental skill in mathematics. It involves finding the set of values that satisfy a given inequality relation. For the function \( f(s) = \sqrt{4s - 5} \), we need 4s - 5 to be non-negative because the square root function is only defined for non-negative numbers.

To solve the inequality \( 4s - 5 \geq 0 \), follow these steps:
  • Add 5 to both sides: \( 4s \geq 5 \)
  • Then, divide both sides by 4: \( s \geq \frac{5}{4} \)
This tells us that the function is defined for all values of \( s \) greater than or equal to \( \frac{5}{4} \). Inequalities can tell us where a function will or won't "work," which in this case means where the expression inside the square root remains valid or non-negative.
Square root function
The square root function, represented as \( f(x) = \sqrt{x} \), is a special type of function that "undoes" squaring.

The most important aspect of the square root function is that it is only defined for non-negative numbers in the real number system. This means you cannot take the square root of a negative number and expect a real result.

  • The symbol \( \sqrt{} \) implies finding a number which, when squared, returns the original number under the root.
  • For example, \( \sqrt{9} = 3 \) because \( 3^2 = 9 \).
  • In contrast, \( \sqrt{-9} \) does not result in a real number.
In our given function \( f(s) = \sqrt{4s - 5} \), the expression \( 4s - 5 \) must be non-negative for the function to be defined. This derivative property heavily influences the domain of the square root functions.
Interval notation
Interval notation is a concise way of expressing the set of numbers that satisfy an inequality. In mathematics, it provides an efficient way to communicate which values are included in or excluded from a solution set.

In the solution to \( f(s) = \sqrt{4s - 5} \), we determined \( s \geq \frac{5}{4} \). Translated to interval notation, this is written as \( \left[\frac{5}{4}, \infty\right) \).

  • The \( [ \) means "including" \( \frac{5}{4} \).
  • The \( \infty \) denotes that there is no upper bound 鈥� \( s \) can be as large as any real number.
  • The \( ) \) following \( \infty \) means infinity itself is not included, as it is not a real number.
This notation is particularly useful in calculus and algebra for quickly and easily determining the domain and range of functions.

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Most popular questions from this chapter

Find the inverse of the given function by using the "undoing process," and then verify that \(\left(f \circ f^{-1}\right)(x)=x\) and \(\left(f^{-1} \circ f\right)(x)=x\). (Objective 4) $$f(x)=-2 x+1$$

If \(y\) is inversely proportional to \(x\), and \(y=\frac{1}{35}\) when \(x=14\), find the value of \(y\) when \(x=16\).

Determine \((f \circ g)(x)\) and \((g \circ f)(x)\) for each pair of functions. Also specify the domain of \((f \circ g)(x)\) and \((g \circ f)(x)\). (Objective 1\()\) \(f(x)=\frac{3}{2 x}\) and \(g(x)=\frac{1}{x+1}\)

Translate each statement of variation into an equation, and use \(k\) as the constant of variation. \(y\) varies directly as the cube of \(x\).

Find the constant of variation for each of the stated conditions. A varies jointly as \(b\) and \(h\), and \(A=72\) when \(b=16\) and \(h=9\).
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