The following function is one-to-one; find its inverse

f(x)=x^3+5

I know that the first step would be to set y = to x so I did y=x^3+5

Then you swap x with y so x=y^3+5.

Then you solve for y and this is where I think i messed up. I did
y^3=x-5 = y=x^3-5

Then the inverse would be f^-1(x)=x^3-5

Is this correct or did I do that last step wrong?

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3. 👁 92
1. Your error is in solving for y in
x = y^3 + 5

y^3 = x-5 , you had that
y = cuberoot(x-5) or (x-5)^(1/3)

check:
let x = 2 in the original y = x^3+5
y = 2^3+5 = 13

now let x = 13 in the invers
y = cuberoot(13-5
= cuberoot(8)
= 2

which is what we started with

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posted by Reiny
2. So what is the inverse?

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posted by Hannah
3. I stated it ....

y = cuberoot(x-5)
or
y = (x-5)^(1/3)
or
f(x) = (x-5)^(1/3)

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posted by Reiny
4. O ok I understand. Thanks!!

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posted by Hannah

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