Answer
See solution
Work Step by Step
For every value of x, there is a different value of y, so this is a one to one function.
As a more formal proof, the derivative of the function is
$f'(x)=\frac{3x^2+2x+1}{2\cdot\sqrt{x^3+x^2+x+1}}$, where $2\cdot\sqrt{x^3+x^2+x+1}\neq0$.
$3x^2+2x+1$ has $D=b^2-4ac=-8$, which is smaller than 0. This means the slope is never equal to 0 and is strictly increasing or decreasing. From the graph, we can see the function is strictly increasing.
