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 2 years ago
Find the Inverse of the function:
f(x)= (x/6)^3  7
My answer was f^1(x) = 6(x+7)^3 is this correct?
 2 years ago
Find the Inverse of the function: f(x)= (x/6)^3  7 My answer was f^1(x) = 6(x+7)^3 is this correct?

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freewilly922
 2 years ago
Best ResponseYou've already chosen the best response.1You exponent is incorrect.

CarlyLiberty
 2 years ago
Best ResponseYou've already chosen the best response.0so would i write it like this? f^1(x) = [6(x+7)]3

CarlyLiberty
 2 years ago
Best ResponseYou've already chosen the best response.0ohhh or is it f^1(x) = 6(x^3 + 7)?

freewilly922
 2 years ago
Best ResponseYou've already chosen the best response.1No, what you did wrong was to assume that \[(y/6)^3 =x+7 \to y/6=(x+7)^3\] That's not how you get rid of exponents. if \[y^4=x\to y=x^{1/4}\]

CarlyLiberty
 2 years ago
Best ResponseYou've already chosen the best response.0okay, so it should be f1(x) = 18(x + 7) then? Thats the only other answer i got...

freewilly922
 2 years ago
Best ResponseYou've already chosen the best response.1You first answer was correct EXCEPT the actual number you put in the exponent. if you had \[y=(x/2)^31\] then to find the inverse swap x and y and solve \[x=(y/2)^31\] then \[x+1=(y/2)^3\] this is where you did something weird. Where I would take the 1/3 root of both sides: \[(x+2)^{1/3} = ((y/2)^3)^{\frac{1}{3}}=y\] you took both sides to the third power.

CarlyLiberty
 2 years ago
Best ResponseYou've already chosen the best response.0Oh okay, i see what i types wrong...thank you so much!
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