2.4 Continuity  (Page 5/16)

Since $f\left(x\right)=x-\text{cos}x$ is continuous over $\left(\text{−}\infty ,\text{+}\infty \right),$ it is continuous over any closed interval of the form $\left[a,b\right].$ If you can find an interval $\left[a,b\right]$ such that $f\left(a\right)$ and $f\left(b\right)$ have opposite signs, you can use the Intermediate Value Theorem to conclude there must be a real number c in $\left(a,b\right)$ that satisfies $f\left(c\right)=0.$ Note that

and

Using the Intermediate Value Theorem, we can see that there must be a real number c in $\left[0,\pi \text{/}2\right]$ that satisfies $f\left(c\right)=0.$ Therefore, $f\left(x\right)=x-\text{cos}x$ has at least one zero.

No. The Intermediate Value Theorem only allows us to conclude that we can find a value between $f\left(0\right)$ and $f\left(2\right);$ it doesn’t allow us to conclude that we can’t find other values. To see this more clearly, consider the function $f\left(x\right)={\left(x-1\right)}^2.$ It satisfies $f\left(0\right)=1>0,f\left(2\right)=1>0,$ and $f\left(1\right)=0.$

No. The function is not continuous over $\left[\mathrm{-1},1\right].$ The Intermediate Value Theorem does not apply here.