Show that $\tan x - 2x$ has a root in $(0,1.4)$.
I got: $f(x) = \tan x - 2x$
$f(0) = 0$, $f(1.4) = -2.775$
Does this mean there is no root in $(0,1.4)$? Because I got $-2.775$ and it did not fall in $(0 , 1.4)$.
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$f(1.4)$ should be positive, while $f(0.1)$ should be negative. – vadim123 Jun 09 '13 at 15:44
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If you put your calculator in radian mode, you'll see that $f$ actually takes on a positive value at $1.4$. Now find an $a$ in the interval where it takes a negative value. Since it is continuous on the closed interval from $0$ to $1.4$, then IVT says there is some $c$ in the interval (specifically, $c$ will be between $a$ and $1.4$) where the function is $0$.
Edit: It's worth noting that the function values found needn't fall in the given interval. For example, we can consider the continuous function $$g(x)=x-3$$ on the interval $[2,4]$. Then $g(2)=-1$ and $g(4)=1,$ so IVT says $g$ has a zero in $(2,4)$, meaning there is some $x\in(2,4)$ such that $g(x)=0$. Of course, $0\notin(2,4)$--in fact, for any $x\in[2,4],$ we have $g(x)\notin[2,4]$--but that has nothing to do with IVT.
Cameron Buie
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1For a bonus point, use the fact that $f(0)=0$ while $f'(0)<0$ to show that $f(a)<0$ for some small positive $a$ – without actually computing such an $a$. – Harald Hanche-Olsen Jun 09 '13 at 15:51