 letumsnemesislh

2022-07-08

How is intermediate value theorem valid for $\mathrm{sin}x$ in $\left[0,\pi \right]$?

It has max value 1 in the interval $\left[0,\pi \right]$ which doesn't lie between values given by $\mathrm{sin}0$ and $\mathrm{sin}\pi$. trantegisis

Expert

Let $f:\left[a,b\right]⟶\mathbb{R}$ be a continuous function and let $y\in \mathbb{R}$. The intermediate value theorem says that if $f\left(a\right)⩾y⩾f\left(b\right)$ or if $f\left(a\right)⩽y⩽f\left(b\right)$, then there is a $c\in \left[a,b\right]$ such that $f\left(c\right)=y$. But it says nothing if $y$ lies outside the interval bounded by $f\left(a\right)$ and $f\left(b\right)$. So, there is no contradiction here. tripes3h

Expert

That's not how it works. The Intermediate Value Theorem says that if $f$ is continuous on [$a$,$b$], then it achieves every value between

and

When $f$ is monotone, it happens that $c,d$ are $f\left(a\right),f\left(b\right)$, but in general it is not the case.

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