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Help!! Let f(x)=7x-13. Find f^-1(x). Let f(x)=x^2-16. Find f^-1(x).

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Do you know how to find an inverse?
Okay thanks..

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I'm still confused..what am I finding the inverse to in my problem?? @hartnn
ok, let y=7x-13 now to find the inverse, isolate x =... ? then just replace 'x' and 'y' (you can replace before isolating also.)
-7x=-13-y?? why would there be a y in there??
and then divide the whole thing by -7, but they're going to be fractions.
y= 7x-13 add 13 to both sides, then divide by 7 [Let there be fractions, its ok]
So basically you set the problem equal to 7x-13, and either make it equal to y or x?? There is no y in the problem? that's why i am confused
x=-13-y/-7, and then do you want me to solve for x on y=7x-13 or that's what y equals?? So you use x to solve for y and y to solve for x??
y = f(x).. They are the same thing
x and y are the same thing??????
i considered f(x) = y for simplicity. i'll write out steps so that you can understand. y= 7x-13 y+13 = 7x x = (y+13)/7 now to find inverse function just replace x and y y= (x+13)/7 so your inverse function is \[f^{-1}(x)=\dfrac{x+13}{7}\] got this ?
f(x) and y are the same thing, so you re-write it as y = 7x-13
got that part?
swop d x for d y.... it becomes x = 7y-13
can you do same thing for f(x)=x^2-16. ?
after swopping, solve for y
7y= x+13 y = (x+13)/7
Yes x=4 + sqrt y
yes, but u have to swop d x and d y back... final answer would be y = 4 + sqrt(x)
did u simplify like this ? \[\sqrt{16+y}=4+\sqrt y \] no, thats incorrect.
by d do you mean the?
sqrt. y+16=x^2 (Root over y+16)
yeah, the inverse function will be \[f^{-1}(x)=\sqrt{x+16}\] no further simplification possible.
ok ? got it ?
Oh got it!!!

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