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• Notes

Show that the equation $\quad \mathbf{e}^{x}=3-2 x$

Has a real solution on $[0,1]$

(1) $e^{x}=3-2 x \rightarrow e^{x}+2 x-3=0 \quad$ let $\quad f(x)=e^{x}+2 x-3$

$f$ is Cont on $R$

$\rightarrow f$ is Cont on $[0,1]$

(2) $f(0)=e^{0}+2(0)-3=1-3=-2<0$

$F(0) < 0 < F(1)$

(3) by IvT, $c \in(0,1)$ such that $f(c)=0$

\ $f$ has a real solution on $[0,1]$

Let $f(x)=x-\ln \left(x^{3}\right)$ . Use the Intermediate Value theorem
to show that $f(x)$ has at least one real root on $[1, e]$

$f(x)=x-\ln \left(x^{3}\right) \quad[1, e]$

(1) $f$ is Cont on $(0, \infty)$

$f$ is cont on $[1, e]$

(2) $f(1)=1- \ln(1)^{3}=1-3 \ln (1)=1$

$\ln (a)^{b}=b \ln (a)$ (+)

$f(e)=e-\ln \left(e^{3}\right)=e-3 \ln (e)=-0.2$ (-)

$f(e) < 0 < f(1)$

(3) by IvT $\Rightarrow c \in(1, e)$

Sueh that $f(c)=0$

\ $f(x)$ has at least one root on $[1, e]$