Given the function f(x) = 4(x+3) – 5, solve for the inverse function when x = 3.
@johnweldon1993 this is my last question, if you don't mind!
Alright, so here...we need the inverse...and then plug in that value x = 3 have you done inverses before?
Yes but they are VERY new to me. This is my first year of Algebra 2.
Ah, well welcome, it only gets harder from here >.< lol no, just kidding :P So For inverses...all we need to do...is switch the positions of the 'x' and the 'y'...and then solve for the 'y' again
So here we have \[\large y = 4(x + 3) - 5\]
Switch the 'x' and the 'y' \[\large x = 4(y + 3) - 5\]
and use algebra to solve for 'y' again... how would we do that?
Combine terms (not sure if that's the proper term) to get 4y+7?
Indeed :) so we have \[\large x = 4y + 7\] so solving that for 'y' we get?
x-4y=7?
Not quite :) so what the whole idea is...is to isolate the 'y' because eventually we want y = ...something right? So where we have \[\large x = 4y + 7\] We want y by itself....so we need to move everything else to the other side of the equation If we subtract 7 from both sides that will isolate the 4y...so that's a start! \[\large x - 7 = 4y\] Now...if we want to solve for y (not 4y) what do we do?
Hm. I'm not sure.
Well... \[\large 4y = x - 7\] Lets see...we have a 4 being MULTIPLIED to the 'y'....so to cancel that...we do what is opposite of multiplication...which is division Since 4 is being multiplied to the y...if we divide that by 4...we will have \[\large \frac{4y}{4} = y\] which is what we want right? we wanted to just have 'y' So that's what we need to do...we need to divide everything by 4 \[\large \frac{x - 7 = 4y}{4}\] \[\large \frac{x - 7}{4} = y\]
Ohh right!
Now, that is the whole process you will ALWAYS use for inverses...so does that make sense? any parts cloudy?
The given answers are single numbers, though.
*We aren't done yet ^_^
OH alright!
Remember we still have the "when x = 3" part to deal with :)
SOOOOOO since we know that \[\large y = \frac{x - 7}{4}\] "when x = 3" what does y equal?
-1?
Indeed! :)
Thank you so very much!
Of course, Anytime! :)
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