It is currently 19 Jul 2019, 10:28

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# In the function above, for what values of x is g(x) a real n

Author Message
TAGS:
Founder
Joined: 18 Apr 2015
Posts: 7386
Followers: 123

Kudos [?]: 1448 [0], given: 6583

In the function above, for what values of x is g(x) a real n [#permalink]  14 May 2019, 01:18
Expert's post
00:00

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
$$g(x)= (2x−3)^{\frac{1}{4}} + 1$$

In the function above, for what values of x is g(x) a real number?

(A) $$x ≥ 0$$

(B) $$x ≥ \frac{1}{2}$$

(C) $$x ≥ \frac{3}{2}$$

(D) $$x ≥ 2$$

(E) $$x ≥ 3$$
[Reveal] Spoiler: OA

_________________
MyGuru Representative
Affiliations: Partner at MyGuru LLC.
Joined: 13 May 2019
Posts: 113
Location: United States
GMAT 1: 770 Q51 V44
GRE 1: Q169 V168
WE: Education (Education)
Followers: 1

Kudos [?]: 98 [1] , given: 2

Re: In the function above, for what values of x is g(x) a real n [#permalink]  14 May 2019, 07:27
1
KUDOS
Expert's post
First remember that raising a quantity to a fractional exponent is equivalent to taking the radical of the number in the exponent's denominator.

Therefore, we can rewrite the expression as g(x) = ∜(2x-3) + 1

Thus, because taking an even root of any negative results in a non-real or imaginary number, we know that 2x - 3 must be ≥ 0.

Solve the inequality by adding 3 to each side and then dividing by 2 to find that the domain of g(x) is x ≥ 3/2, which is choice C.
_________________

Stefan Maisnier

Re: In the function above, for what values of x is g(x) a real n   [#permalink] 14 May 2019, 07:27
Display posts from previous: Sort by