Intermediate Value Theorem

Tags

Cegep1

Mathematics

Word count

281 words

Reading time

2 minutes

Abbr. I. V. T.

Suppose that $f$ is continuous over $[a,b]$ where $f(a)\ne f(b)$.
Let $N$ be any number between $f(a)$ and $f(b)$,
then $\mathrm{\exists }c\in [a,b]$ such that $f(c)=N$.

To effectively use I. V. T. for functions that don't pass the horizontal line test, you may need to cut the interval.

[!example]+ Show that $\sin x+1=\sqrt{x}$ is solvable.

$$\begin{array}{rl}\sin x+1& =\sqrt{x}\\ \sin x+1-\sqrt{x}& =0\end{array}$$

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

$$\begin{array}{rlr}f(0)& =\sin 0+1-\sqrt{0}& =1>0\\ f(\pi )& =\sin \pi +1-\sqrt{\pi }& =1-\sqrt{\pi }<0\end{array}$$

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

[!example]+ Show that $\cos x=x^2$ is solvable for $x\in (0,1)$.

$$\begin{array}{rl}\cos x& =x^2\\ \cos x-x^2& =0\end{array}$$

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

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

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

Contributors

Ajitani Hifumi

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34fe6ed-Vault update: 2024-09-27T16:12:31