Show that the function $g(x)=x^4+x^3+1$ is one-to-one on $[0,2]$.
My attempt
To prove one-to-oneness, we shall use the definition, that is, if $f(x_1)=f(x_2)$ , then $x_1=x_2$ for all $x_1,x_2\in[0,2]$
Suppose $f(x_1)=f(x_2)$, then ${x_1}^4+{x_1}^3+1={x_2}^4+{x_2}^3+1$
Which wasn't much of an attempt, as I got stuck.
Also I am rather new to proving conjections and surjections.
Help! Thanks in advance!