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.

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

Prep Toolkit

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My Rewards