Inverse of Function f: R -> R Defined by f(x) = x^3+1

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SUMMARY

The function f: R -> R defined by f(x) = x^3 + 1 is confirmed to be one-to-one, as demonstrated by the condition that f(a) = f(b) implies a = b. The inverse of this function is accurately calculated as f-1(x) = (x - 1)1/3. Additionally, the composition of the function with itself, denoted as (f o f), results in f(f(x)) = (x3 + 1)3 + 1. The discussion clarifies the definitions of one-to-one and onto functions, ensuring a comprehensive understanding of the topic.

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Homework Statement



f: R -> R defined by f(x) = x^3 + 1

a) determine if it is one to one
b)find its inverse
c) calculate (f o f)



Homework Equations



if f(x) is one to one, if a = b, f(a) = f(b)


The Attempt at a Solution



a) yes it is because a^3 + 1 can't equal to b^3 + 1 unless a = b, by definition, this function is onto
b)the inverse is f^-1 (x) = (x - 1)^(1/3)
c) f o f = f(f(x) = f(x^3 + 1) = (x^3 + 1)^3 + 1

just want to make sure i got this, thanks :)
 
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For #1, by definition the function is 1-1 since you just said that if f(a) = f(b) then a = b, not onto. Onto means that if you have something in your codomain, b in R, then there's an element in your domain, a in R, such that f(a) = b for every b in R in the mapping F: R -> R.

For #2, let's see. If you have f(x) = x^3 + 1, then using basic manipulations, you get (y-1)^(1/3) = x if you let y = f(x). Then, change this to f^-1(x) = (x-1)^(1/3), which is what you got.

For #3, you got it. In this case, your x is equal to x^3+1 in F(x). Therefore, if F(x) = x^3 +1, then F(x^3 + 1) = (x^3 + 1)^3 + 1, which is what you also got.

Re-write #1 and all is good.
 
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