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 $f(x) = 0$?
 A: 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.
A: No, as the real root in this case is not a nice number.  Instead, you can use the fact that polynomials (such as $x^3-x-1$) are continuous to help you show that there is a root.
If you can find some $x$ such that $x^3-x-1<0$, and some $y$ such that $y^2-y-1>0$, then there must be some $z$ between $x$ and $y$ such that $z^3-z-1=0$.

(source: wolframalpha.com)
A: Pick $x=1$ and $x=2$. $f(1)=-1$ and $f(2)= 5$.  As $f$ is continious, then it has to cut $x-axis$ somewhere between $x\in(1,2)$, that is it has at least one root.  
A: 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.
