Determine if the function $ f: ℝ → ℝ, f(x)= (1+x) / (3x-1)$ is:
a. Injective
b Surjective
c. Bijective
My answer:
a. $ f(x_{1}) = f(x_{2}) $
$ (1+x_{1})/(3x_{1} -1) = (1+x_{2})/(3x_{2} -1) $
$ (x_{1})/(3x_{1} -1) = (x_{2})/(3x_{2} -1) $
$ (x_{1})/(3x_{1}) = (x_{2})/(3x_{2}) $
$ (x_{1})/(x_{1}) = (x_{2})/(x_{2}) $
$ x_{1} = x_{2} $
Injective: Yes
b. $ f(x) = (1+x)/(3x-1) $
$ y = (1+x)/(3x-1) $
$ y(3x-1) = 1 + x $
$ y(3x) - y = 1 + x $
$ y(3x) - x = 1 + y $
$ 3x(y-1) = 1+ y $
$ 3x = (1+ y)/(y-1) $
Surjective: No
c. Bijective: No (Not both injective and surjective)
Is this correct?
As said in the comment: $\mathbb R$ cannot serve as domain of the function.
You should go for a function $\mathbb R-\{\frac13\}\to\mathbb R$ because the function is not defined for $x=\frac13$.
Observe that: $$\frac{1+x}{3x-1}=\frac13+\frac43\frac1{3x-1}$$ and that the second term cannot take value $0$ so that the function cannot take value $\frac13$. This shows that the function is not surjective.
Further to prove injectivity it is enough to prove injectivity on $\mathbb R-\{\frac13\}$ of the more simple function: $$x\mapsto\frac1{3x-1}$$