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Chapter 1: Problem 84

Find the domain of the function.$$f(x)=\frac{\sqrt{x-5}}{x-7}$$

Short Answer

Expert verified
The domain of the function \(f(x) = \frac{\sqrt{x-5}}{x-7}\) is given by \(x \geq 5, x \neq 7\). Alternative notation: \([5,7) \cup (7, \infty)\)

Step by step solution

01

Constraint of the square root function

To have a real output, the radicand (the expression inside the square root) must be greater than or equal to zero. Hence we first set up the inequality \(x - 5 \geq 0\). Solving this gives \(x \geq 5\). This represents all the values of \(x\) that will make the square root part of the function real.
02

Constraint of the fractional function

For the fractional part of the function, we need to ensure that the denominator is not equal to zero as this would make the function undefined. Hence, we set up the equation \(x - 7 \neq 0\). Solving this gives \(x \neq 7\). This means all real numbers except 7 are valid for the denominator.
03

Combine the constraints to determine the domain

By combining the results from Steps 1 and 2, we can see that all values \(x \geq 5\) are allowable provided \(x \neq 7\), because the denominator also cannot equal zero. Therefore, the domain of \(f(x)\) is \(x \geq 5, x \neq 7\). This can also be written as \([5,7) \cup (7, \infty)\) in interval notation.

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Key Concepts

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

Square Root Function
In mathematics, a square root function involves finding the number that, when multiplied by itself, gives the original number underneath the root symbol. For example, the square root of 9 is 3 because \(3 \times 3 = 9\).
When dealing with square root functions like \(f(x) = \sqrt{x-5}\), the expression under the root, called the "radicand," must be non-negative (greater than or equal to zero). Why? Because square roots of negative numbers are not real numbers in basic math.

Let's dive into our exercise:
  • The radicand here is \(x-5\).
  • To ensure we have real outputs, we set \(x-5 \geq 0\).
  • Solving gives \(x \geq 5\), meaning any \(x\) value from 5 onward is suitable for this function part.
This result tells us where the square root function is valid to calculate real outputs.
Fractions in Functions
Fractions in functions introduce a critical aspect: the denominator cannot be zero, as division by zero is undefined.
When dealing with fractions like \(\frac{\sqrt{x-5}}{x-7}\), it's crucial to determine what makes the denominator zero and exclude those values from the domain.

Here's how it works in our example:
  • The denominator is \(x-7\).
  • Setting \(x-7 eq 0\) ensures it doesn't equal zero.
  • Solving gives \(x eq 7\), meaning that \(x\) cannot be 7 to keep the function valid.
In sum, the denominator constraint ensures the fraction stays valid and produces real numbers.
Interval Notation
Interval notation is a concise way of describing sets of numbers, often used to express domains of functions.
It uses brackets and parentheses to show whether endpoints are included or not in the set.

Within our problem, after combining the constraints from the square root and fraction, we find:
  • The values of \(x\) must be 5 or greater (due to the square root).
  • But \(x\) cannot be 7 (due to the fraction).
This gives us the domain in set terms: all numbers from 5 to just before 7, plus all numbers greater than 7.
In interval notation, we write this combined domain as \[ [5, 7) \cup (7, \infty) \]. This notation efficiently shows the function's domain without confusion.

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Most popular questions from this chapter

Use the functions \(f(x)=\frac{1}{8} x-3\) and \(g(x)=x^{3}\) to find the indicated value or function. $$\left(f^{-1} \circ f^{-1}\right)(-6)$$

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