If f(x) = 9x + 1, find f^–1 (f (3)).
f(x) is a linear function so it is one-to-one throughout its entire domain
this means that \[\large f^{-1}(f(x)) = x\] is true for all values of x
So how would i solve it?
FInd inverse of y=9x+1 once you do then plug in 3. You know how to do inverse?
example: plug in x = 17 into the function to get some output f(17) you then plug the output f(17) into the inverse function and you'll get back 17 as the final result
y=9x+1 swap y and x then solve for y :D x=9y+1 1st step of swapping
so \[\large f^{-1}(f(17)) = 17\]
yeah i know how to do the inverse (:
but what do i do after i find the inverse?
what did you get for the inverse
y= x+1 over 9.... and then i plug in 3?
after finding inverse plug in your f(3)
it should be \[\large f^{-1}(x) = \frac{x-1}{9}\]
what you do is plug in x = 3 into f(x) first once you get your answer, you plug that result into the inverse function and evaluate
oh oops thats what i meant (:
Ohhhhhh okay got it thanks (:
yw
so 3 is what i got is that right?
3 is correct, if you looked back at my example, I got \[\large f^{-1}(f(17)) = 17\] and it turns out that \[\large f^{-1}(f(x)) = x\] is true for all x (so just replace x with 3 and you're done) this works out because f(x) is a linear function
ohhh so I can do that will all of them?
yeah if you know f(x) is a linear function, then \[\large f^{-1}(f(x)) = x\] can be used
okay thanks (:
you're welcome
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