If f(x) = g(x - 1), where g(x) = |x| + 1, which of the following must

Question

If f(x) = g(x - 1), where g(x) = |x| + 1, which of the following must be true? (A) f(x) < 0 (B) f(x) = 0 (C) f(x) > 0 (D) f(x) = 2 (E) f(x) = g(x) Functions.png

Bunuel · Accepted Answer

Since g(x) = |x| + 1, then f(x) = g(x - 1) = |x - 1| + 1 = (nonnegative value) + 1 = positive. Thus, f(x) > 0.

Answer: C.

kpop1234567890 · Answer

Bunnel soln is perfect but i did it longer way
If f(x) = g(x-1), where g(x) = |x| + 1, which of the following must be true?

Now
g(x-1)=|x-1|+1
If x-1>0 (implies x>1)
g(x-1)=x-1+1=x Always positive

if x-1<0 (implies x<1)
g(x-1) = -(x-1)+1
g(x-1) = 2 - x
This will again always be positive because all values of x<1

Hence f(x)=g(x-1)>0

f(x)>0

c

Kinshook · Answer

callumdye

Asked: If f(x) = g(x-1), where g(x) = |x| + 1, which of the following must be true?

f(x) = |x-1| + 1 > 0 Since |x-1|>=0

IMO C

BrushMyQuant · Answer

↧↧↧ Detailed Video Solution to the Problem ↧↧↧

https://www.youtube.com/watch?v=CudIHYnZa6M

Given that f(x) = g(x - 1) and g(x) = |x| + 1 and we need to find which of the following must be true?

To find g(x-1) we need to compare what is inside ( ) in g(x) = |x| + 1 and g(x-1)

=> We need to substitute x with x-1 in g(x) = |x| + 1 to get the value of g(x-1)

=> g(x-1) = |x-1| + 1
 => f(x) = g(x-1) = |x-1| + 1

Now, f(x) = Modulus of some number + 1, and we know that modulus of any number is non-negative
=> f(x) > 0 + 1
=> f(x) > 1
=> f(x) > 0

So,  Answer will be C 
Hope it helps!

Watch the following video to MASTER Functions and Custom Characters

https://www.youtube.com/watch?v=b7JoMX0gk8M

KarishmaB · Answer

Functions may look hard but there are just a few things you need to keep in mind while working with them. Check this post for those: https://anaprep.com/algebra-functions/ Given: g(x) = |x| + 1 Then g(x-1) = |x-1| + 1 (wherever there is x, simply put x-1) Given: f(x) = g(x - 1) = |x-1| + 1 Now there is no one value that f(x) will take. It can take infinite values for different values of x. It can be 0, it can be 2 and it can be many other values. But note that for no value of x will f(x) be negative. It is a sum of two non-negative quantities. Answer (C)

DanTheGMATMan · Answer

Very common motif on the GMAT focus: just extract some logical conclusion from an algebraic setup. The absolute value has to be nonnegative, so f(x) will be greater than or equal to 1:
 https://www.youtube.com/watch?v=oGRXqFQeBeA

NEYR0N · Answer

why are we not taking scenarios where (x-1)>=0 , (x-1) < 0 ?

Bunuel · Answer

There's no need to split into cases like (x-1) ≥ 0 or (x-1) < 0 because:

|x-1| is  always  ≥ 0 - that's the definition of absolute value.

So, |x-1| + 1 is  always  ≥ 1.

Thus, f(x) = |x-1| + 1 is  always  > 0 - no matter what x is.

That’s why option (C) is correct  without  needing to check cases.

NEYR0N · Answer

I'm surely not understanding something. What is the crux here ? https://gmatclub.com/forum/on-the-numbe ... 16667.html How are we assessing critical values 3 and 4 here, but not in this question? Thanks,

Bunuel · Answer

Two questions are totally different.

In the question at hand, we are assessing |x - 1| + 1. How does it matter what x is or what x - 1 is? Who cares? The point the absolute value of x - 1, so |x - 1|, is ALWAYS non-negative. So |x - 1| + 1 will ALWAYS be positive - irrespective of x!

ManishaPrepMinds · Answer

Rewrite f(x) using g’s definition
f(x) = g(x − 1) = |x − 1| + 1
Use the property of absolute value
For any real x, |x − 1| ≥ 0 (absolute value is a distance, never negative)
f(x) = |x − 1| + 1 ≥ 0 + 1, so f(x) is always strictly positive
-> f(x) > 0

GSV · Answer

f(x) = g(x - 1) = |x - 1| + 1 = (positive value or 0) + 1 = positive. Thus, f(x) > 0.

Hence answer is C.

Usernamevisible · Answer

Bunuel  is this right approach -

Since g(x) = |x| + 1,
and |x| ≥ 0 for all x,
g(x) ≥ 1.
Therefore, g(x) is always positive.
-----------------------------------

Since f(x) = g(x - 1),
and x - 1 can be any real number,
f(x) is also always positive.
-----------------------------

Therefore,
f(x) > 0.
Answer: (C)
-----------

Key idea:
x - 1 is just another input. If g is positive for every possible input, then g(x - 1) must also be positive for every possible x. This is sufficient justification for the answer.

egmat · Answer

Hi Usernamevisible,

Your conclusion is right, and your core insight (g is never less than  1 , for  any  input) is exactly the property that settles this question. The answer is  C .

There's just one link worth tightening so the justification is airtight rather than intuitive.

Your sentence  "x - 1 is just another input, so g(x - 1) must also be positive"  is true - but it's leaning on intuition. The clean way to make it rigorous is to actually  form f(x) by substitution , the same move you'd make for any function:

- g(x) = |x| + 1
- To get g(x - 1), replace every x with  x - 1 : g(x - 1) =  |x - 1| + 1 
- So f(x) =  |x - 1| + 1

Now the positivity isn't an appeal to "g is always positive" in the abstract - you can  see  it directly: |x - 1| >=  0  for every real x, so f(x) = |x - 1| + 1 >=  1 . That's strictly greater than 0, no matter what x is. So  f(x) > 0  must be true.

Why this matters

"The input doesn't change the output's sign" is the right instinct, but on the GMAT you want to land it on the actual expression. Once you write f(x) = |x - 1| + 1, every other option falls away cleanly: it's never 0 (min is 1), never a fixed  2  (that needs |x - 1| = 1), and not equal to g(x) in general.

So your answer is correct, and with the substitution written out, your justification is fully sufficient too.

Answer: C

If f(x) = g(x - 1), where g(x) = |x| + 1, which of the following must

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If f(x) = g(x - 1), where g(x) = |x| + 1, which of the following must

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