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I am reading this example in the book for Pre-Calculus and it is explaining how functions are shifted left or right using g(x)=f(x-1). Here is what it says in the book.

Define a function g by

g(x) = f(x-1)

where f is the function defined by f(x)=x^2 , with the domain of f the interval [-1,1].

a)Find the domain of g. b)Find the range of g.

Solution: a) The formula defining g shows that g(x) is defined precisely when f(x-1) is defined, which means that x-1 must be in the interval [-1,1], which means that x must be in the interval [0,2]. Thus the domain of g is the interval [0,2].

Okay, so I am having a hard time understanding the solution from the book. Particularly the part when they say "which means that x-1 must be in the interval [-1,1], which means that x must be in the interval [0,2]." If x-1 is in the interval [-1,1], shouldn't the domain of g be [-2,0]? Since I plug in -1,0,1 into x-1. Could someone explain why the domain would be [0,2] instead of [-2,0]?

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Related: math.stackexchange.com/questions/133185/… –  Austin Mohr Sep 9 '12 at 3:59

1 Answer 1

up vote 1 down vote accepted

If f's domain is the interval [-1,1], that means that whatever you plug into f has to be in the interval [-1,1]. Because you are plugging (x-1) into f, (x-1) has to be in [-1,1].

If (x-1) is in [-1,1], that means that x has to be in [0,2]. To see why this is, you could write this out as:

x-1 ≥ -1

x-1 ≤ 1

Then just solve for x:

x ≥ 0

x ≤ 2

Because x must be in [0,2], the domain of g is [0,2].

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I kind of understand it now, but why did you set up the inequalities? Since it's a closed interval, wouldn't you use the greater than or equal to and less than or equal to signs? –  Kot Sep 9 '12 at 4:00
    
Yes, you're absolutely right. That was a mistake on my part, and I fixed it. –  Brandon Pickering Sep 9 '12 at 4:13
    
Thank you for teaching me the inequality technique! I understand how it works now :D. –  Kot Sep 9 '12 at 4:28

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