/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 58 Determine the domain of the func... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Determine the domain of the function according to the usual convention. $$f(x)=\sqrt{-x}+\frac{2}{x+1}$$

Short Answer

Expert verified
Answer: The domain of the function is \((-∞, -1) \cup (-1, 0]\).

Step by step solution

01

Identify the restrictions from the square root

In the function \(f(x) = \sqrt{-x}\), we need the expression under the square root to be non-negative (i.e. greater than or equal to 0) since taking the square root of a negative number would result in a complex number, and we are looking for a real-valued function. Therefore, we have the inequality: $$ -x \geq 0 $$
02

Solve the inequality for the square root restriction

Now, we need to solve this inequality to find which values of x satisfy the restriction. To solve the inequality, we can multiply both sides by -1, remembering to flip the inequality sign when multiplying by a negative number: $$ x \leq 0 $$
03

Identify the restrictions from the division

Next, we need to consider the division part of the function, \(\frac{2}{x+1}\). We should ensure that the denominator is never equal to 0, as this would result in an undefined value (division by zero). So, we must find the values of x for which the denominator is not 0: $$ x+1 \neq 0 $$
04

Solve the inequality for the division restriction

To find the values of x that satisfy this restriction, we need to solve the equation: $$ x \neq -1 $$
05

Combine both restrictions to find the domain

Finally, we need to combine both restrictions to find the domain of the function. Since x must be less than or equal to 0 according to the first restriction, and x must not be equal to -1 according to the second restriction, the domain of the function includes all x-values less than or equal to 0, except for -1 itself. We can express the domain in interval notation as follows: $$ (-\infty, -1) \cup (-1, 0] $$

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Square Root Restrictions
When working with square roots, it's essential to ensure that the expression inside the square root is non-negative. In mathematical terms, if you encounter a term like \( \sqrt{-x} \), the part inside the square root, \(-x\), must be \( \geq 0 \). This is because square roots of negative numbers result in complex numbers, which are outside the scope of real-valued functions we typically deal with in many elementary contexts.

To address this, we set up the inequality \(-x \geq 0\), which we need to solve for \(x\). This involves multiplying both sides by \(-1\), remembering a crucial rule: always flip the inequality symbol when multiplying or dividing by a negative number. Therefore, \(-x \geq 0\) becomes \(x \leq 0\). This result tells us that \(x\) can be any value less than or equal to zero for the square root to remain valid and real.

This concept is pivotal because square root expressions dictate a direct constraint on the domain of any function containing them. Always ensure values under the root are non-negative to maintain function validity.
Division by Zero
In mathematics, division by zero is undefined. This is because dividing any number by zero does not yield a finite value. As a critical rule for any function involving division, ensure the denominator cannot be zero. In an expression like \( \frac{2}{x+1} \), this means identifying where the denominator \( x+1 \) equals zero.

Practically, this is executed by solving \( x+1 eq 0 \). To find out where this denominator becomes zero, we solve \( x+1 = 0 \), which simplifies to \( x = -1 \). Therefore, \( x \) cannot be \(-1\), as attempting division by zero at this point would lead the function to be undefined.

Understanding division by zero restrictions is crucial to defining a function's domain because it highlights values that must be excluded to ensure the function is well-defined and operates over a real number set. Always verify that the denominator avoids zero to keep functions operable and valid.
Inequalities in Domain Restriction
Combining restrictions derived from square roots and division is key to finding the domain of a composite function. Each component of the function contributes its own set of restrictions, and these need to be considered together.

For a function like \( f(x) = \sqrt{-x} + \frac{2}{x+1} \), first, ensure \( x \leq 0 \) from the square root restriction and \( x eq -1 \) from the division restriction. Each of these contributes a different condition to the overall domain.

To find the combined domain, express the valid range of \( x \) values meeting both restrictions. We interpret \( x \leq 0 \), while excluding \( x = -1 \), resulting in an interval notation: \((-\infty, -1) \cup (-1, 0]\). This means \( x \) can take any real value less than or equal to zero, except \(-1\).

Understanding and applying inequality-based restrictions aids in accurately determining permissible values for your variables, hence defining the function's domain comprehensively and correctly.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In each part, compute \(g(a), g(b),\) and \(g(a b),\) and determine whether the satement " \(g(a b)=g(a) \cdot g(b)\) " is true or false for the given function. (a) \(g(x)=x^{3}\) (b) \(g(x)=5 x\) (c) \(g(x)=-2\)

Example \(11(b)\) showed how we create a table of values for a function when you get to choose all the values of the inputs. The technique presented does not work for Casio calculators. This exercise is designed for users of Casio calculators. \(\cdot \quad\) Enter an equation such as \(y=x^{3}-2 x+3\) in the equation memory. This can be done by selecting TABLE in the MAIN menu. \- Return to the MAIN menu and select LIST. Enter the numbers at which you want to evaluate the function as List 1 \(\cdot \quad\) Return to the MAIN menu and select TABLE. Then press SET-UP [that is, 2nd MENU] and select LIST as the Variable; on the LIST menu, choose List 1. Press EXIT and then press TABL to produce the table. \- Use the up/down arrow key to scroll through the table. If you change an entry in the X column, the corresponding \(y_{1}\) value will automatically change. (a) Use this technique to duplicate the table in Example \(11(\mathrm{b})\) (b) Change the number -11 to \(10,\) and confirm that you've obtained \(10^{3}-2(10)+3\)

Use algebra to find the inverse of the given one-to-one function. $$f(x)=-3 x+5$$

Determine the domain of the function according to the usual convention. $$h(x)=\frac{\sqrt{x-1}}{x^{2}-1}$$

If \(f(x)=\left\\{\begin{array}{ll}x^{2}+2 x & \text { if } x<2 \\ 3 x-5 & \text { if } 2 \leq x \leq 20\end{array}\right.\) find (a) The domain of \(f\) (b) \(f(-3)\) (c) \(f(-1)\) (d) \(f(2)\) (e) \(f(7 / 3)\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.