$[-/6$ Points$]$

EXAMPLE $10$

Show that there is a solution of the equation

$7x^3-9x^2+3x-2=0$

between $1$ and $2.$

SOLUTION

Let $f(x)=7x^3-9x^2+3x-2=0.$ We are looking for a solution of the given equation, that is$,$ a number $c$ between $1$ and $2$ such that $f(c)=q,.$ Therefore we take $a=1,b=,$ and $N=q,n$ the Intermediate Value Theorem. We have

$f(1)=7-9+3-2=-1<mn>0</mn>$

and $f(2)=56-36+6-2=24>0$

Thus, $N=0f(1)f(2)fcf(c)=7x^3-9x^2+3x-2=0cf(1\mathrm{.}1)=-0\mathrm{.}273<mn>0</mn>$ and $f(1\mathrm{.}2)=0\mathrm{.}736>0f(1\mathrm{.}13)=-0\mathrm{.}001821<mn>0</mn>$ and $f(1\mathrm{.}14)=0\mathrm{.}094408>0f(1)<mn>0</mn>$; that $isN=0is$ a number between $f(1)$ and $f(2).$ Now $fis$ continuous since $itis$ a polynomial, $so$ that the Intermediate Value Theorem says there $is$ a number $c$ between $1$ and

$$ such that $f(c)=.In$ other words, the equation $7x^3-9x^2+3x-2=0$ has $at$ least one solution $cin$ the open interval, $in$ interval notation, $.$

$In$ fact, $we$ can locate a solution more precisely $by$ using the Intermediate Value Theorem again. Since

$f(1\mathrm{.}1)=-0\mathrm{.}273<mn>0</mn>$ and $f(1\mathrm{.}2)=0\mathrm{.}736>0$

a solution must lie between $(smaller)$ and $(larger).$ A calculator gives, $by$ trial and error,

$f(1\mathrm{.}13)=-0\mathrm{.}001821<mn>0</mn>$ and $f(1\mathrm{.}14)=0\mathrm{.}094408>0$

Based $on$ these inequalities, a solution lies $in$ the open interval, $in$ interval notation, $$

Need Help? $$