State the domain of y=-2f(3(x+1))-5Hello, This is an example from my online class and I'm having trouble understanding a few things here. Here is the example word for word.   If F=f(x) has the...

State the domain of y=-2f(3(x+1))-5

Hello,

This is an example from my online class and I'm having trouble understanding a few things here.

Here is the example word for word.

 

If F=f(x) has the domain {x|-2<x<4,x ER}, state the domain of y=-2f(3(x+1))-5.

 

The domain of y=f(x) is {x|-2<x<4,x ER}.

After the horizontal compression by a factor of 1/3 and the translation (horizontal) of 1 unit left:

{x|(-2/3)-1<x(4/3)-1,xER} --> ={x|(-5/3)<x<(1/3),xER}

Therefore, the domain of y=-2f(3(x+1))-5 is {x|(-5/3)<x<(1/3),xER}

 

My problem is I don't understand how {x|(-2/3)-1<x(4/3)-1,xER} become {x|(-5/3)<x<(1/3),xER}? How did x|(-2/3)-1 become x|-5/3??

 

I've attached a link to the original example if you want to view it to gain a better understanding of this.

http://i184.photobucket.com/albums/x275/maythany/mathquestion.png

 

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted on

Given a function f(x) with the domain `{x|-2<x<4,x in RR}` state the domain of `y=-2f(3(x+1))-5`

Note the transformations on f(x): the -2 performs a vertical stretch of factor 2 and a reflection over the horizontal axis. Neither of these transformations affects the domain. (They affect the range.)

The 3 performs a horizontal compression of factor 1/3. Thus the original domain is shrunk by a factor of 1/3 to `-2/3<x<4/3`

The 1 performs a horizontal translation (shift) left 1 unit. This shifts the domain 1 unit to the left. The new domain `-2/3-1<x<4/3-1`

** But `-2/3-1=-2/3-3/3=-5/3` and `4/3-1=4/3-3/3=1/3` so we can write the new domain as `-5/3<x<1/3` **

The -5 performs a vertical translation of 5 units down -- this only affects the range, not the domain.

Therefore, the new domain is `{x|-5/3<x<1/3,x in RR}`

 

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