The domain of the function f(x) = \frac{\sqrt{x - 1{x+1} is the

Question

The domain of the function f(x) = \frac{\sqrt{x - 1} }{x+1} is the set of all real numbers that are A. greater than 1 B. greater than or equal to 1 C. not equal to -1 D. less than or equal to 1 E. less than 1 https://gmatclub.com/forum/download/file.php?id=81451 GMAT-Club-Forum-gdfjkydu.png

gmatophobia · Accepted Answer

We have to ensure the below constraints are met -

1) The value under the square root should always be non-negative.

\sqrt{x - 1}

x - 1 \geq 0

x \geq 1

2) The value of the denominator should not be zero

x + 1 
eq 0

x 
eq -1

Combining information from both the constraints

x \geq 1

Option B

JeffTargetTestPrep · Answer

The domain of a function is the set of all possible values of a variable for which every expression in the formula of the function is defined.

f(x) = √(x – 1)/(x + 1)

A square root expression is defined only if the expression under the square root is non-negative:

x – 1 ≥ 0

x ≥ 1

A fraction is defined only if its denominator isn’t equal to zero.

x + 1 ≠ 0

x ≠ -1

If x ≥ 1, the condition x ≠ -1 is clearly satisfied as well, so the domain of the function is:

x ≥ 1

Answer: B

Paras96 · Answer

f(x) =  /x+1

The domain is defined as the value of x that can be used in the function so that f(x) remains real,

here, f(x) has 2 parts => In the numerator,  , to be real (x-1) should be greater than or equal to zero which gives, x E  E

Taking final values, we get x E  Hence B

RitwikAgrawal · Answer

>30 sec solution

Solve for denominator: X not equal to 1 (dived by zero is undefined)
Solve for numerator: X greater than equal to one (root of negative is not real)

DanTheGMATMan · Answer

Tests the 2 forbidden outcomes:

https://www.youtube.com/watch?v=CEyXzzEVn50

Ilanchezhiyan · Answer

Sir , could you please explain how x not equal to -1 and x>=1; can be combined together as x>=1, what if x=-2, it could still be a real number?

Bunuel · Answer

The condition x ≥ 1 already excludes x = -1 because -1 is less than 1. There’s no need to state x ≠ -1 separately since it’s automatically not included in x ≥ 1.

Archit3110 · Answer

Plug in values.. real number is target
Square root cannot be -ve c,d,e are out.

Option A does not consider 1..
Option B is correct as least value is 0 ..

seb11 · Answer

Because of the denominator you cannot have x=-1 
Because of the nominator being square root of x-1 (which has to be positive) you can put in values that make (x-1) equal or above 1.

So the domain is the intersection of:
all numbers excluding -1
and
greater than or equal to 1.

So the answer is b

kingbucky · Answer

Let’s find the domain of the function  f(x) = \frac{\sqrt{x - 1}}{x + 1} , where the square root applies only to the numerator.

The domain is the set of all real numbers  x  for which the function is defined. 
For the numerator  \sqrt{x - 1}  to be defined, we need  x - 1 \geq 0 \implies x \geq 1 . 
The denominator  x + 1 
eq 0 \implies x 
eq -1 . 
Combining these,  x \geq 1  already excludes  x = -1  since  -1 < 1 , so the domain is   . This matches the option "greater than or equal to 1."

Answer:  greater than or equal to 1

shishimaru · Answer

To tackle problems like finding the domain of a function such as  f(x) = \frac{\sqrt{x - 1}}{x + 1} , a solid understanding of a few key mathematical concepts is essential. Let’s break down the background knowledge that will help you approach these types of questions systematically.

First, you need to understand the concept of the domain of a function. The domain is the set of all possible input values  x  for which the function is defined, meaning the output  f(x)  must be a real number. For a function to be defined, you must check for restrictions imposed by operations like division, square roots, logarithms, or other operations that have specific requirements.

One critical operation in this problem is the square root. The square root function  \sqrt{y}  is defined for real numbers only when  y \geq 0 . So, for an expression like  \sqrt{x - 1} , you need  x - 1 \geq 0 , which simplifies to  x \geq 1 . This restriction often appears in functions involving square roots, and you’ll frequently solve inequalities to determine where the expression inside the square root is non-negative.

Another key operation here is division, since the function has a denominator  x + 1 . Division by zero is undefined in mathematics, so you must ensure the denominator is not zero. For  x + 1 , this means  x + 1 
eq 0 \implies x 
eq -1 . This is a common restriction for rational functions (functions that are ratios of two expressions). Always identify points where the denominator is zero and exclude them from the domain.

When combining multiple restrictions, you take the intersection of the conditions. In this problem, the square root requires  x \geq 1 , and the denominator requires  x 
eq -1 . Since  x \geq 1  means  x  is at least 1, it automatically excludes  x = -1  (because  -1 < 1 ). The domain is the set of  x  values that satisfy all conditions simultaneously, which here is   .

Finally, break the function into parts: the numerator  \sqrt{x - 1}  has its own domain, and the entire fraction  \frac{	ext{numerator}}{x + 1}  adds the denominator’s restriction.

With this background—understanding domains, square root and division restrictions, solving inequalities, using interval notation, and breaking down composite functions—you can confidently solve similar problems by identifying restrictions and combining them logically.

supremekai · Answer

Focus on key math concepts to find the domain of functions like  f(x) = \frac{\sqrt{x - 1}}{x + 1} , which is the set of all  x  values where the function gives a real number output.

First, the square root  \sqrt{x - 1}  requires the inside to be non-negative:  x - 1 \geq 0 \implies x \geq 1 . 
Second, the denominator  x + 1  cannot be zero, so  x + 1 
eq 0 \implies x 
eq -1 . Since  x \geq 1  already excludes  x = -1 , the domain is  x \geq 1 .

Know these basics: square roots need  	ext{expression} \geq 0 , division means the denominator can’t be zero, and combine conditions by taking the overlap.

lordbeerus · Answer

Cheatsheet: Points to Remember for Finding the Domain of Functions

When finding the domain of a function (all  x  values where the function gives a real number), look for operations that make the function undefined. Here’s what to check:

1. Division by Zero (Fractions) 
 If there’s a denominator, like in  \frac{1}{x - 2} , it can’t be zero. Solve  	ext{denominator} = 0  and exclude those  x  values. Example:  x - 2 = 0 \implies x 
eq 2 .

2. Square Roots (Even Roots) 
 For  \sqrt{	ext{expression}}  or even roots like  \sqrt {	ext{expression}} , the expression inside must be non-negative:  	ext{expression} \geq 0 . Solve this. Example:  \sqrt{x + 3}  needs  x + 3 \geq 0 \implies x \geq -3 .

3. Rational Inequalities (e.g., Inside a Root) 
 If a fraction is inside a root, like  \sqrt{\frac{x - 1}{x + 2}} , make the expression fit the root’s rule:  \frac{x - 1}{x + 2} \geq 0 . Find where the numerator and denominator are zero ( x - 1 = 0 \implies x = 1 ,  x + 2 = 0 \implies x = -2 ), test values between these points, and exclude where the denominator is zero.

4. Combining Multiple Restrictions 
 If the function has multiple parts, like  \frac{\sqrt{x - 1}}{x + 2} , check each part: 
 -  \sqrt{x - 1} :  x \geq 1 . 
 - Denominator  x + 2 
eq 0 \implies x 
eq -2 . 
 The domain is where all conditions overlap.

5. No Restrictions Means All Real Numbers 
 If there are no restricting operations, like in  f(x) = x^2 + 3 , the domain is all real numbers.

6. Quick Tips 
 - Break the function into parts and check each for restrictions. 
 - Solve inequalities step-by-step for roots. 
 - Always exclude points where the function is undefined (like division by zero). 
 - Express the domain as inequalities, like  x \geq 1  or  x 
eq 2 .

maingi95 · Answer

What am I missing here the domain has to be real but no where in the question it mentions the Range has to be real, Why are values below 1 (except -1) not included in the domain?

Bunuel · Answer

If x is less than 1, the expression under the square root is negative, which is undefined in GMAT since it only deals with real numbers.

Su1206 · Answer

I encountered this question in a mock. I correctly selected B. The official answer shows A. Its official mock. i am confused.

The domain of the function f(x) = \frac{\sqrt{x - 1{x+1} is the

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The domain of the function f(x) = \frac{\sqrt{x - 1{x+1} is the

Prep Toolkit

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