PLEASE HELP!! If f^-1(x)=4/3x+8, what is f(x)??
so the inverse of a function sends the x to y and y to x, so the inverse of the inverse, does what?
Replace f^-1 by \(x\) Replace x by \(f(x)\) then, solve for \(f(x)\)
@FibonacciChick666 switch the variables
yep, so f^-1(x) let's call that y', we can call the x for the inverse function x'. Now since we swap the x and the y just to get the inverse function, can you tell me what y'= in terms of the original function?
If that doesn't make sense tell me, I'll try to explain better.
y is = 4/3x+8
ok, so y'=4/3x'+8
so that we can use x and y for f(x)
is the answer is f(x)=3/4x-8
That's possible, but I have not done it yet
oh okay
so for me personally, to avoid annoying variable confusion. I name the inverse g(x). So g(x)=x for f(x) and the x in g(x)=f(x) or y in the original f(x)
so rewriting in terms of f(x) \[g(x):=~~~x=\frac{4}{3}y+8\]
so now I solve for y. Then I will be done
so can you see if using that you get the same answer? I think it may be different
ok, I know I'm being confusing so, to get the inverse you swap x and y right? well to get the inverse of the inverse, ie the original functions, you swap the variables again.
yea i know
ok so solve: \[ x=\frac{4}{3}y+8\] for y and see if you get the same answer
( you shouldn't)
show your steps, I think I know where you are going wrong
\[x=\frac{ 4 }{ 3 }y+8\]
\[-3x=4y+8\]
why do you have a minus 3x?
\[f(x)=\frac{ 3 }{ 4}x-8\]
that is an incorrect step
because i put 3 outside the = and change to -3
help me
yea... that's not how it works
ok so \[x=\frac{4}{3}y+8\] You have to reverse the order of operations to solve for y. So if you use PEMDAS, the reverse is SADMEP. So we start by undoing any addition or subtraction. Do we have any?
can you show mw the 2nd step?? maybe i might figure it out
well, first answer, is there any addition going on?
yes
so how do we undo addition?
what is the opposite of addition?
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