Intermediate Value Theorem

Examples

Show that $\sin x + 1 = \sqrt{x}$ is solvable.

\begin{aligned} \sin x + 1 & = \sqrt{x} \\ \sin x + 1 - \sqrt{x} & = 0 \end{aligned}

Let $f (x) = \sin x + 1 - \sqrt{x}$ .
$f (x)$ is continuous $\forall x \in [0, \infty)$ .
Consider $x \in [0, π]$ :

\begin{aligned} f (0) & = \sin 0 + 1 - \sqrt{0} & = 1 > 0 \\ f (π) & = \sin π + 1 - \sqrt{π} & = 1 - \sqrt{π} < 0 \end{aligned}

Let $N = 0 \in [f (π), f (0)]$ ,
so by I. V. T., $\exists c \in [0, π]$ such that $f (c) = N = 0$ .

Show that $\cos x = x^{2}$ is solvable for $x \in (0, 1)$ .

\begin{aligned} \cos x & = x^{2} \\ \cos x - x^{2} & = 0 \end{aligned}

Let $f (x) = \cos x - x^{2}$ .
$f (x)$ is continuous $\forall x \in R$ .
Consider $x \in [0, 1]$ , then

\begin{aligned} f (0) & = \cos 0 - 0^{2} = 1 \\ f (1) & = \cos 1 - 1^{2} < 0 & , since \cos 1 \in (- 1, 1) . \end{aligned}

Let $N = 0 \in [f (1), f (0)]$ ,
so by I. V. T., $\exists c \in [0, 1]$ such that $f (c) = N = 0$ .

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