If f(x) = x + x^2/(3x) + 4, which of the following numbers CANT be in

Question

If  f(x) = x + \frac{x^2}{3x} + 4 , which of the following numbers can't be in the domain of f(f(x))?

I. 3
II 0
III -3

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

https://gmatclub.com/forum/download/file.php?id=80515

ShilpiAgnihotrii · Answer

@ Bunuel   KarishmaB  @ chetan2u , experts can u pls help me solving this question...I am in major doubt with this question. Thanks in advance!

Bunuel · Answer

Please check whether f(x) = (x + x^2)/(3x + 4) is correct and formatted properly (brackets and so on). Thank you.

ShilpiAgnihotrii · Answer

Edited the format

Bunuel · Answer

You need to post a screenshot of the question.

Bunuel · Answer

And the OA given in the source is A?

ShilpiAgnihotrii · Answer

Its E Bunuel, i guess by mistake i have marked A

AnishPatel · Answer

The correct answer is indeed A. 1 ONLY.
Here's why:
Step 1: Simplify f(x)
First, simplify f(x) = x + 4 + (x^2/3x) = x + 4 + (x/3).

Step 2: Find f(f(x))
Now, substitute f(x) into f(x) to get f(f(x)).

Step 3: Analyze the Domain
The domain of f(f(x)) is restricted where the denominator of the fraction in f(x) is zero, which occurs when 3x = 0 --> x = 0.

However, when we plug x = 3 into f(x), we don't get a zero denominator. Instead, 
we get: f(3) = 3 + 4 + (3/3) = 8

But when we plug x = 3 into f(f(x)), 
we get: f(f(3)) = f(8) = 8 + 4 + (8/3*8)

Now, the issue arises when we try to plug x = 3 into the inner function f(x) of f(f(x)). 
We get: f(3) = 3 + 4 + (3/3*3) = 3 + 4 + 3 = 10

Then, when we plug f(3) = 10 into the outer function f(x), 
we get: f(f(3)) = f(10) = 10 + 4 + (10/3*10)

This is where the problem occurs. When x = 3, the inner function f(x) evaluates to 10, but then the outer function f(x) tries to divide by 3*10, which is not zero. However, when we simplify f(x), we get x + 4 + (x/3), and when x = 3, we get 3 + 4 + (3/3) = 8. Then, when we plug x = 8 into the outer function f(x), we don't get a division by zero error.

However, there is another issue. When we plug x = 3 into the outer function f(x) of f(f(x)), 
we get: f(f(3)) = f(8) = 8 + 4 + (8/3*8)

But what if we plug x = 3 directly into the outer function f(x) of f(f(x))? 
We get: f(3) = 3 + 4 + (3/3) = 8

Then, when we plug x = 3 into the outer function f(x), 
we get: f(3) = 3 + 4 + (3/3*3)

Now, we see the problem. When x = 3, the outer function f(x) tries to divide by 3*3, which is zero.

Therefore, the correct answer is indeed A. 1 ONLY.

ManifestDreamMBA · Answer

Given we have x in the denominator, we know 0 is out of domain
So eliminate A,C,D.

Let's check for -3
 f(-3) = -3 + \frac{(-3)^2}{3(-3)} + 4 
 f(-3) = -3 + \frac{9}{(-9)} + 4 
 f(-3) = -3 + -1 + 4 
 f(-3) = 0

Now if I solve for f(f(x)) = f(0), we will have 0 in the denominator.

So E is the answer

KarishmaB · Answer

Since x is in the denominator, it cannot be 0. So f(f(x)) is not defined when x = 0. Hence II needs to be a part of the solution. 
Possible answers: B, D, E

Can x be 3? 
f(x) = f(3) = 7
f(f(3)) = f(7) = 7 + 7/3 + 4 - All ok.

Can x be -3?
f(x) = f(-3) = 0
f(f(-3)) = f(0) - Not defined because 0 cannot be in the denominator.

Hence answer is that 0 and -3 are not in domain of f(f(x)).

Answer (E)

If f(x) = x + x^2/(3x) + 4, which of the following numbers CANT be in

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If f(x) = x + x^2/(3x) + 4, which of the following numbers CANT be in

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