Question

Determine whether the function f(x) = x^3 - 1 from Z to Z is one to one

Answer 1

\[Df:f(x)\rightarrow \mathbb{R} \]
with \[ x_{1},x_{2}\in Df\]
Let\[ f(x _{1})=f(x_{2}) \]
\[<=> x _{1}^{3}-1=x_{2}^{3}-1 <=> x_{1}^{3}=x_{2}^{3} <=> x_{1}^{3}-x_{2}^{3} \]
\[<=> (x_{1}-x_{2})(x_{1}^{2}+x_{1}x_{2}+x_{2}^{2}=0 <=>(i) x_{1}=x_{2} (1) \]
and \[ (ii)x_{1}^{2}+x_{1}x_{2}+x_{2}^{2}=0 <=>2x_{1}^{2}+2x_{1}x_{2}+2x_{2}^{2}=0 <=>\] \[(x_{1}+x_{2})^{2}+x_{1}^{2}+x_{2}^{2}=0 <=>x_{1}=x_{2}=0\] that it's included in (1)
so, \[ f(x_{1})=f(x_{2})\]
So the function is "1-1"

Answer 2

This can also be done graphically - the graph of f(x) = x^3 - 1 is shown below:



From the graph, there exists only 1 values of y for every value of x, where x, y ∈ R.