I am not interested in finding roots but interested in proving that the function has real roots.
Suppose a function $f(x) = x^2 – 1$
This function obviously has real roots.
$x = {-1, 1}$
How could I prove this without actually finding the roots?
Trial and error could work, number theory even? (modulus etc?) Calculus, any methods?
Thanks!
Best Answer
Since $f(0)=-1<0$ and $f(2)=3>0$, it is clear that $f(x)=0$ for some $x\in[0,2]$ by the intermediate value theorem.
Similarly, you can see that there is a root of $f$ for some $x\in[-2,0]$.
In general, it is hard to find zeroes for an arbitrary function, even if it is continuous. The intermediate value theorem doesn't work for functions that only touch the $y=0$ axis, for example for $f(x)=x^2$.