/*! 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 27 Find the domain of each function... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the domain of each function \(n(x)=\sqrt{(x-3)(x+2)^{2}}\)

Short Answer

Expert verified
The domain of \(n(x) = \sqrt{(x-3)(x+2)^2}\) is \([-2, 3] \cup (3, \infty)\)."}

Step by step solution

01

Understand the Function

The function given is \( n(x) = \sqrt{(x-3)(x+2)^2} \). Since it's a square root function, for it to be defined, the expression inside the square root must be non-negative.
02

Set the Expression to be Non-Negative

The expression under the square root is \((x-3)(x+2)^2\). We need \((x-3)(x+2)^2 \geq 0\). This means that the product must be zero or positive.
03

Analyze the Expression

The expression \((x+2)^2\) is always non-negative because it is a square. Hence, it will be zero only when \(x = -2\). The critical points to analyze the sign of \((x-3)\) are \(x = 3\) and \(x = -2\).
04

Test Intervals Between Critical Points

Divide the real number line into intervals based on critical points: \((-\infty, -2), (-2, 3), (3, \infty)\). Test points from each interval to determine where the product is non-negative.
05

Test Interval \((-\infty, -2)\)

Pick a test point, for example, \(x = -3\). Then \((-3-3)((-3+2)^2) = (-6)(1) = -6\), which is negative. So, \(n(x)\) is not defined in \((-\infty, -2)\).
06

Test Interval \((-2, 3)\)

Pick a test point, for example, \(x = 0\). Then \((0-3)((0+2)^2) = (-3)(4) = -12\), which is negative. So, \(n(x)\) is not defined in \((-2, 3)\).
07

Test Interval \((3, \infty)\)

Pick a test point, for example, \(x = 4\). Then \((4-3)((4+2)^2) = (1)(36) = 36\), which is positive. So, \(n(x)\) is defined in \((3, \infty)\).
08

Check Values at Critical Points

Check \(x = -2\): \((-2-3)((-2+2)^2) = (-5)(0) = 0\), which is non-negative, so \(n(x)\) is defined at \(x = -2\). Check \(x = 3\): \((3-3)((3+2)^2) = (0)(25) = 0\), so \(n(x)\) is defined at \(x = 3\).
09

Conclusion About the Domain

Combining all the intervals and points where the function is defined, the domain of \(n(x)\) is \([-2, 3] \cup (3, \infty)\).

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.

Interval Notation
Interval notation is a mathematical way of describing a set of numbers along a number line. It's a compact alternative to listing numbers individually. Instead, you mention just the endpoints of the interval, making it easy to communicate which numbers belong to the set.
  • When an interval includes an endpoint, it uses brackets like "]-2, 3]", which means both -2 and 3 are included in the set.
  • If the endpoint is not included, parentheses are used, such as "(3, \infty)", meaning all numbers greater than 3 are included, but 3 itself is not.
  • The infinity symbol, \(\infty\), implies stretching towards the endless in either direction.
In the solution provided, the domain of the function was shown to be \

One App. One Place for Learning.

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

Get started for free

Study anywhere. Anytime. Across all devices.