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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 $$[1, e]$$
No comments yet
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 $$[1, e]$$
No comments yet
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