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We know that IVP of a continuous function says that if $f:\mathbb R\rightarrow \mathbb R$ be a continuous function on $\mathbb R$ then between $[a, b]$ there will be at least one real root of $f(x)=0$ if $f(a)f(b)<0$ OR either even number of roots or no root of $f(x)=0$ if $f(a)f(b)>0$.

Suppose we consider $f(x)=|\sin x|$ in $[\frac{\pi}{2}, \frac{3\pi}{2}]$.

It is continuous function on $[\frac{\pi}{2}, \frac{3\pi}{2}]$ and $f(\frac{\pi}{2})f(\frac{3\pi}{2})>0$.

The problem is: there is exactly only one real root of $f(x)=0$ in $[\frac{\pi}{2}, \frac{3\pi}{2}]$ which is $\pi$. viz $f(x)=0$ for $x=\pi$.

Is this violating IVP?

KON3
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    What do you mean by IVP? – Crostul Sep 15 '14 at 08:50
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    I have never seen the IVP (Intermediate Value Property?) stated like this. I guess your version is true for polynomials, where roots are counted with multiplicity. – Taladris Sep 15 '14 at 08:51
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    The intermediate value property states that for if $f(a)f(b)<0$, then there exists at least one $x\in(a,b)$ such that $f(x)=0$. Your version is clearly wrong as your example shows. Another more straight-forward counterexample is $f(x)=|x|$. – J.R. Sep 15 '14 at 08:54
  • ya. I have got my answer. Thanks to all of you – KON3 Sep 15 '14 at 08:55

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That is not the initial value property. It states that if a function is continuous in the domain (a, b) and the values of f at a and b be f(a) and f(b) then then f takes all the values between f(a) and f(b).

Your statement about roots is wrong.

For example consider a parabola $y^2=4x$ in the interval (-1, 1).

Your statement about odd roots is cool but about even roots is wrong since the functions can touch the x-axis only once as shown by the parabola.

Jasser
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