Which of the following functions has as its domain the set of all real

Question

Which of the following functions has as its domain the set of all real numbers?

I.   f(x)=(64x)^\frac{1}{3}

II.   f(x)=\sqrt{2x^2-8x+8}

III.  f(x)=\frac{2}{2x^2-1}

A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III

anish777 · Answer

I think the OA should be changed. What do you think ? the answer seems to be D.

Bunuel · Answer

Yes,   f(x)=\frac{2}{2x^2-1}  won't be defined if the denominator, 2x^2 - 1 is 0, so when  x = \frac{1}{\sqrt{2}}  or  x = -\frac{1}{\sqrt{2}} . Edited the OA. Thank you!

LamboWalker · Answer

Official Explaination:-

The domain of a function f(x) is all values of x for which f(x) is well-defined.

I. Since all odd roots of any real number are well-defined,  (64x)^\frac{1}{3}  is   well-defined for all real values of x.

II.  \sqrt{2x^2-8x+8}=\sqrt{2(x^2-4x+4)}=\sqrt{2(x-2)^2}=(x-2)\sqrt{2} , which is   well-defined for all real values of x.

III. If  x=\frac{1}{\sqrt{}2} , then the denominator  (2x^2-1)  would be 0. Thus,  \frac{2}{2x^2-1}  is   NOT well-defined for all real values of x.

D is the correct answer choice.

DrAnkita91 · Answer

I. f(x)=(64x)^1/3
In this if we take f(x)= -1
then it leads to -64^1/3, which is not a real number right?
please correct me

Bunuel · Answer

No. Even roots from negative numbers are not real numbers. However, odd roots from negative numbers are. For example,  \sqrt {-64}=-4 .

SatvikVedala · Answer

For   f(x)=(64x)^\frac{1}{3}

Consider x = -1 => f(x) =  (-64)^\frac{1}{3}  => f(x) = -2
Consider x = 1 => f(x) =  (64)^\frac{1}{3}  => f(x) = 2

Consider x = -2 => f(x) =  (-64*2)^\frac{1}{3}  => f(x) = -5.0396
 (-2)^\frac{1}{3}  =  (-1)^\frac{1}{3}  *  (2)^\frac{1}{3}  = -1.2599

Consider x = 2 => f(x) =  (64*2)^\frac{1}{3}  => f(x) = 5.0396

Likewise odd roots of x will be real numbers. Hence the domain ranges all real numbers

ItzSam · Answer

How is option 2, well defined for all real values of x ? what if the value of X is 2, then in that case the solution would be 0. Kindly explain this part. And what exactly do we mean, when we say - "well-defined for all real values of X"

Bunuel · Answer

If a function produces a numerical result for a specific value, then it's defined for that value. For example, if f(x) = 1/x, this function will be undefined for x = 0 because 1/0 is undefined. Essentially:

• For a fraction, the function is undefined for values that make the denominator 0.
• For even roots, the function is undefined for values that make the expression under the root negative, as even roots on the GMAT are defined only for non-negative values (zero and positive).

For   f(x)=\sqrt{2x^2-8x+8} , it is defined for x = 0 because we get   f(0)=\sqrt{0}=0 . The square root of 0 is valid, so there's no issue.

kayarat600 · Answer

Official explanation doesn't quite make sense to me anyone that can help not sure this question is even relevant?

Krunaal · Answer

We are asked to determine which of the given functions will stay valid / are well defined, when x is any real number. For 1. and 2. If you substitute any real number in place of x, there will be a valid f(x) For 3. There is a case where f(x) can be invalid when denominator becomes 0 due to a possible real value of x That's what the above solutions in the thread explain, please highlight if you have any specific doubt. https://blog.targettestprep.com/gmat-fu ... a-Function => theory on domain and range of a function. Hope it helps.

ManishaPrepMinds · Answer

I: f(x) = (64x)^(1/3) 
This is a cube root function.
Cube roots are defined for all real numbers, unlike square roots.
 Domain: All real numbers

II: f(x) = √(2x^2 - 8x + 8) 
This is a square root function requiring the expression inside to be non-negative.
2x^2 - 8x + 8 ≥ 0
2x^2 - 8x + 8 = 2(x^2 - 4x + 4) = 2(x - 2)^2
(x - 2)^2 ≥ 0 for all real x (perfect square is always non-negative)
 Domain: All real numbers

III: f(x) = 2/(2x^2 - 1) 
This is a rational function that's undefined when the denominator equals zero.
2x^2 - 1 ≠ 0
 Finding excluded values:  2x2 - 1 = 0 
2x^2 = 1, x^2 = 1/2 
x = ±√(1/2) 
 Domain: All real numbers except x = ±√(1/2)

Answer: D. I and II only

Which of the following functions has as its domain the set of all real

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Which of the following functions has as its domain the set of all real

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