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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 $$[1, e]$$

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