Inactive Tutor answered • 03/11/21
We can think about this in terms of transformations -- shifts, stretches, shrinks, etc.
In this case, f(4x+2) represents 2 horizontal transforms applied to the graph, meaning the range stays as it is, f(x) ε (6 , 8].
We should rewrite the function with the 4 factored out to put it in standard form: f(4(x+1/2))
This expression indicates the graph will first be shrunk by a factor of 4, ie the points will move 4 times closer to the y-axis than they were, or the x-coordinates will be divided by 4. Then, the points are shifted left by 1/2, meaning we subtract 1/2 from the x-coordinates.
Applying these two transforms gives x ε (- 3 , 1] , which is the domain of the new function.
You can verify that applying 4x + 2 to these new endpoints gives the old endpoints of the domain interval.
Inactive Tutor
Happy to help -- looks like you may have had several questions on this same topic, and you can obviously apply the same concepts to those questions as well. Remember however that the transforms for y are the opposite of those for x. Eg 4f(x) + 2 represents a vertical stretch by 4, followed by a shift up 2 (in contrast to the horizontal shrink and shift left for the question I answered above).03/11/21
Samuel B.
What if the function was 6 f(x) + 4? How would you go about solving this?03/13/23
Inactive Tutor
That’s a vertical stretch by a factor of 6 followed by a shift up by 4. Both are vertical transforms, meaning the domain stays the same this time, (-10,6], whereas the endpts of the range get multiplied by 6, then increased by 4, becoming (40,52].03/13/23
Peter L.
Thanks a lot, this is really helpful03/11/21