Prove that $f(x)=x^3 −x−1$ has at least one real root. How would I go about proving this?
Would I try finding a value for $x$ that will make $f(x) = 0$?
It's continuous, odd powered polynomial, which means it goes from $-\infty$ to $+\infty$, which in turn means that it crosses $x$ axis somewhere in between.
So, $\space$the$\space$ $f(x) = x^3 - x - 1$

Then,$\space$ $x$ $(x^2 - 1) - 1$, by pulling out an $x$.

$\rightarrow$ $x - 1$ is equal to $x = 1$

$\rightarrow$ $x^2 - 1$ is equal to $x^2 = 1$, then sqrt both sides, $x = 1$

However, If you apply the limit test of $lim_x\rightarrow_\infty$ of the $f(x) = x^3 -x -1$, then the $f(x) =\infty$ so the function is continuous.