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...
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??
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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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