Using the Bolzano–Cauchy Theorem, prove that the equation $x^2+x−\cos x=0$ has at least one solution in the interval [0,1].
Let $f : [a,b] \to \mathbb R$ be a continuous function if the signs of $f(a)$ and $f(b)$ are different (for example, $f(a) > 0, f(b) < 0$). Then there is a point $c \in [a, b]$ such that $f(c) = 0$.
For $x=0 \implies 0^2+0−\cos 0=-1$
For $x=1 \implies 1^2+1−\cos 1 = 2 - \cos 1 \approx 2 - 0.54\ldots \approx 1.46\ldots$
Since $f(a) < 0$ and $f(b) > 0$ we have $f(a) f(b) < 0$, so by the theorem there is a $c \in [a ,b]$ such that $f(c) = 0$.
This is enough? seems very simple, am i missing something? Because the theorem doesn't say how to find the root, it just guarantees that there is at least one.
Thanks.