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Question-133341




Question Number 133341 by mohammad17 last updated on 21/Feb/21
Answered by mathmax by abdo last updated on 21/Feb/21
1)f(x)=x^3  +4x^2 −10    ⇒f^′ (x)=3x^2  +8x >0 on[1,2] ⇒f is increazing  on [1,2] we have f(1)=5−10=−5<0 and f(2)=8+16−10=14>0 ⇒  ∃! x_0 ∈]1,2[ /f(x_0 )=0  we can use newton method  x_(n+1) =x_n −((f(x_n ))/(f^′ (x_n )))  if we take x_0 =(3/2)=1,5 ⇒  x_1 =x_0 −((f(x_0 ))/(f^′ (x_0 ))) and x_2 =x_1 −((f(x_1 ))/(f^′ (x_1 ))) .....
$$\left.\mathrm{1}\right)\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{3}} \:+\mathrm{4x}^{\mathrm{2}} −\mathrm{10}\:\:\:\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{x}\right)=\mathrm{3x}^{\mathrm{2}} \:+\mathrm{8x}\:>\mathrm{0}\:\mathrm{on}\left[\mathrm{1},\mathrm{2}\right]\:\Rightarrow\mathrm{f}\:\mathrm{is}\:\mathrm{increazing} \\ $$$$\mathrm{on}\:\left[\mathrm{1},\mathrm{2}\right]\:\mathrm{we}\:\mathrm{have}\:\mathrm{f}\left(\mathrm{1}\right)=\mathrm{5}−\mathrm{10}=−\mathrm{5}<\mathrm{0}\:\mathrm{and}\:\mathrm{f}\left(\mathrm{2}\right)=\mathrm{8}+\mathrm{16}−\mathrm{10}=\mathrm{14}>\mathrm{0}\:\Rightarrow \\ $$$$\left.\exists!\:\mathrm{x}_{\mathrm{0}} \in\right]\mathrm{1},\mathrm{2}\left[\:/\mathrm{f}\left(\mathrm{x}_{\mathrm{0}} \right)=\mathrm{0}\:\:\mathrm{we}\:\mathrm{can}\:\mathrm{use}\:\mathrm{newton}\:\mathrm{method}\right. \\ $$$$\mathrm{x}_{\mathrm{n}+\mathrm{1}} =\mathrm{x}_{\mathrm{n}} −\frac{\mathrm{f}\left(\mathrm{x}_{\mathrm{n}} \right)}{\mathrm{f}^{'} \left(\mathrm{x}_{\mathrm{n}} \right)}\:\:\mathrm{if}\:\mathrm{we}\:\mathrm{take}\:\mathrm{x}_{\mathrm{0}} =\frac{\mathrm{3}}{\mathrm{2}}=\mathrm{1},\mathrm{5}\:\Rightarrow \\ $$$$\mathrm{x}_{\mathrm{1}} =\mathrm{x}_{\mathrm{0}} −\frac{\mathrm{f}\left(\mathrm{x}_{\mathrm{0}} \right)}{\mathrm{f}^{'} \left(\mathrm{x}_{\mathrm{0}} \right)}\:\mathrm{and}\:\mathrm{x}_{\mathrm{2}} =\mathrm{x}_{\mathrm{1}} −\frac{\mathrm{f}\left(\mathrm{x}_{\mathrm{1}} \right)}{\mathrm{f}^{'} \left(\mathrm{x}_{\mathrm{1}} \right)}\:….. \\ $$
Answered by mathmax by abdo last updated on 21/Feb/21
2)we can contnuity of f(x)=(√(((10)/x)−4x))  the limit verify f(L)=L ⇒L=(√(((10)/L)−4L)) ⇒L^2 =((10)/L)−4L ⇒  L^3 =10+4L^2  ⇒L^3  −4L^2 −10 =0 after we solve x^3 −4x^2 −10=0  ....
$$\left.\mathrm{2}\right)\mathrm{we}\:\mathrm{can}\:\mathrm{contnuity}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\frac{\mathrm{10}}{\mathrm{x}}−\mathrm{4x}} \\ $$$$\mathrm{the}\:\mathrm{limit}\:\mathrm{verify}\:\mathrm{f}\left(\mathrm{L}\right)=\mathrm{L}\:\Rightarrow\mathrm{L}=\sqrt{\frac{\mathrm{10}}{\mathrm{L}}−\mathrm{4L}}\:\Rightarrow\mathrm{L}^{\mathrm{2}} =\frac{\mathrm{10}}{\mathrm{L}}−\mathrm{4L}\:\Rightarrow \\ $$$$\mathrm{L}^{\mathrm{3}} =\mathrm{10}+\mathrm{4L}^{\mathrm{2}} \:\Rightarrow\mathrm{L}^{\mathrm{3}} \:−\mathrm{4L}^{\mathrm{2}} −\mathrm{10}\:=\mathrm{0}\:\mathrm{after}\:\mathrm{we}\:\mathrm{solve}\:\mathrm{x}^{\mathrm{3}} −\mathrm{4x}^{\mathrm{2}} −\mathrm{10}=\mathrm{0} \\ $$$$…. \\ $$

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