Help with a question that is on a exam

[gusrohar]

Hi! Unsure whether or not this is the right place to ask this question:

If f(x) = √(x+1)² - √(x-1)²

(a) f(x) = 2; (b) f(x) = 2x; (c) f(x) = 2√x; (d) none of (a)-(c).

My first instict is to remove the squares and the roots so that f(x) = x+1 - (x-1) which in turn gives me the answer (a). This is incorrect. I am guessing it has to do with the fact that x is not defined to be ≥ 0. Any help?

 

[Deleted member 4993]

Hi! Unsure whether or not this is the right place to ask this question:

If f(x) = √(x+1)² - √(x-1)²

(a) f(x) = 2; (b) f(x) = 2x; (c) f(x) = 2√x; (d) none of (a)-(c).

My first instict is to remove the squares and the roots so that f(x) = x+1 - (x-1) which in turn gives me the answer (a). This is incorrect. I am guessing it has to do with the fact that x is not defined to be ≥ 0. Any help?

Domain of f(x) is x≥1

(a) and (b) has unrestricted domain - so the answer is (d). This is a "sneaky" question.

 

[Quaid]

f(x) = √(x+1)² - √(x-1)²

Hello gusrohar:

The typing above looks ambiguous.

Are the binomials squared before taking the square root, or are the radicals themselves being squared?

Ciao

 

[HallsofIvy]

(I thought I had answered this before. Did you put it on another math forum?)

Assuming you mean \(\displaystyle \sqrt{(x+1)^2}- \sqrt{(x-1)^2}\) then, as you have been told, that is NOT equivalent to "x+ 1- (x- 1)= 2" but is equivalent to \(\displaystyle |x+1|- |x- 1|\).

Now, since absolute value depends upon whether the argument is positive or negative, look at three cases:

1) x< -1 where both x+1= x- (-1) and x- 1 are negative so |x+1|= -(x+1) and |x-1|= -(x- 1).

2) \(\displaystyle -1\le x< 1\) where x+ 1 is non-negative but x- 1 is still negative so |x+1|= x+ 1 and |x- 1|= -(x-1).

3) \(\displaystyle 1\le x\) where both x+ 1 and x- 1 are positive so |x+1|= x+ 1 and |x- 1|= x- 1.