Question Number 190260 by alcohol last updated on 30/Mar/23
![f : [1, 3] →R , f(x) = (1/x) A(1, 1) B(1, (1/3)) B′(b, (1/b)) , b ≥ 1 Find i. equation of line AB′ ii. equation of tangent T ′ to C_f at point with x = ((1 + b)/2) iii. Study relative positions of L_(AB ′) , T ′ to C_f](https://www.tinkutara.com/question/Q190260.png)
$${f}\::\:\left[\mathrm{1},\:\mathrm{3}\right]\:\rightarrow\mathbb{R}\:,\:{f}\left({x}\right)\:=\:\frac{\mathrm{1}}{{x}} \\ $$$${A}\left(\mathrm{1},\:\mathrm{1}\right) \\ $$$${B}\left(\mathrm{1},\:\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$${B}'\left({b},\:\frac{\mathrm{1}}{{b}}\right)\:,\:{b}\:\geqslant\:\mathrm{1} \\ $$$${Find} \\ $$$${i}.\:{equation}\:{of}\:{line}\:{AB}' \\ $$$${ii}.\:{equation}\:{of}\:{tangent}\:{T}\:'\:{to}\:{C}_{{f}} \:{at}\:{point} \\ $$$${with}\:{x}\:=\:\frac{\mathrm{1}\:+\:{b}}{\mathrm{2}} \\ $$$${iii}.\:{Study}\:{relative}\:{positions}\:{of}\:{L}_{{AB}\:'} \:,\:{T}\:'\:{to}\:{C}_{{f}} \\ $$
Answered by a.lgnaoui last updated on 31/Mar/23
![i.equation of line AB^′ AB^′ :portion of droite AB^′ (y_1 =a^′ x+b^′ ) verifie { ((a^′ +b^′ =1)),((ba^′ +b′=(1/b))) :} y_1 =((−x)/b)+(1/b)+1 ii)equation of tengente T^: to C_f at point:x_0 =((1+b)/2) (df_((x)) /dx)=((−1)/x^2 ) x_0 =((1+b)/2) ⇒ f^′ (x)′=((−4)/((1+b)^2 )) ((f(x)−f(x_0 ))/(x−x_0 ))=((−4)/((1+b)^2 )) y=((4(((1+b)/2)−x))/((1+b)^2 ))+(2/(1+b))=((2(1+b−2x+1+b))/((1+b)^2 )) y_2 =((−4)/((1+b)^2 ))x+(4/(1+b)) iii)Γ^′ is betwen tengente AB′ and C_f ((1/b)>(1/((1+b)^2 ))) with b>1 y_1 −y_2 =(((−x)/b)+((b+1)/b))+((4/((1+b)^2 ))x−(4/(1+b))) =((4/((1+b)^2 ))−(1/b))x+(((b+1)^2 −4b)/(b(1+b))) =((−(b−1)^2 )/(b(1+b)^2 ))x+(((b−1)^2 )/(b(1+b))) (((b−1)^2 )/(b(1+b)))[((−x)/(b+1))+1] forme a^(′′) x+b^(′′) (b^(′′) >0 a^(′′) <0) ⇒ T^′ ′ ander Γ^′](https://www.tinkutara.com/question/Q190316.png)
$${i}.{equation}\:{of}\:{line}\:{AB}^{'} \\ $$$${AB}^{'} :{portion}\:{of}\:{droite}\:{AB}^{'} \left({y}_{\mathrm{1}} ={a}^{'} {x}+{b}^{'} \right) \\ $$$${verifie}\:\:\begin{cases}{{a}^{'} +{b}^{'} =\mathrm{1}}\\{{ba}^{'} +{b}'=\frac{\mathrm{1}}{{b}}}\end{cases}\:\:\: \\ $$$$\:\:\:\:\:\boldsymbol{{y}}_{\mathrm{1}} =\frac{−\boldsymbol{{x}}}{\boldsymbol{{b}}}+\frac{\mathrm{1}}{\boldsymbol{{b}}}+\mathrm{1} \\ $$$$\left.\boldsymbol{{ii}}\right){equation}\:{of}\:{tengente}\:{T}^{:} \:{to}\:{C}_{{f}} \:{at}\: \\ $$$${point}:{x}_{\mathrm{0}} =\frac{\mathrm{1}+{b}}{\mathrm{2}} \\ $$$$\:\frac{{df}_{\left({x}\right)} }{{dx}}=\frac{−\mathrm{1}}{{x}^{\mathrm{2}} }\:\:\:{x}_{\mathrm{0}} =\frac{\mathrm{1}+{b}}{\mathrm{2}}\:\:\Rightarrow\:{f}^{'} \left({x}\right)'=\frac{−\mathrm{4}}{\left(\mathrm{1}+{b}\right)^{\mathrm{2}} }\: \\ $$$$\frac{{f}\left({x}\right)−{f}\left({x}_{\mathrm{0}} \right)}{{x}−{x}_{\mathrm{0}} }=\frac{−\mathrm{4}}{\left(\mathrm{1}+{b}\right)^{\mathrm{2}} } \\ $$$${y}=\frac{\mathrm{4}\left(\frac{\mathrm{1}+{b}}{\mathrm{2}}−{x}\right)}{\left(\mathrm{1}+{b}\right)^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{1}+{b}}=\frac{\mathrm{2}\left(\mathrm{1}+{b}−\mathrm{2}{x}+\mathrm{1}+{b}\right)}{\left(\mathrm{1}+{b}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\boldsymbol{{y}}_{\mathrm{2}} =\frac{−\mathrm{4}}{\left(\mathrm{1}+\boldsymbol{{b}}\right)^{\mathrm{2}} }\boldsymbol{{x}}+\frac{\mathrm{4}}{\mathrm{1}+\boldsymbol{{b}}} \\ $$$$\left.{iii}\right)\Gamma^{'} {is}\:{betwen}\:\:{tengente}\:{AB}'\:{and}\:{C}_{{f}} \\ $$$$\left(\frac{\mathrm{1}}{{b}}>\frac{\mathrm{1}}{\left(\mathrm{1}+{b}\right)^{\mathrm{2}} }\right)\:\:\:{with}\:\:{b}>\mathrm{1} \\ $$$$\:\:\:\:{y}_{\mathrm{1}} −{y}_{\mathrm{2}} =\left(\frac{−{x}}{{b}}+\frac{{b}+\mathrm{1}}{{b}}\right)+\left(\frac{\mathrm{4}}{\left(\mathrm{1}+{b}\right)^{\mathrm{2}} }{x}−\frac{\mathrm{4}}{\mathrm{1}+{b}}\right) \\ $$$$=\left(\frac{\mathrm{4}}{\left(\mathrm{1}+{b}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{{b}}\right){x}+\frac{\left({b}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}{b}}{{b}\left(\mathrm{1}+{b}\right)} \\ $$$$=\frac{−\left({b}−\mathrm{1}\right)^{\mathrm{2}} }{{b}\left(\mathrm{1}+{b}\right)^{\mathrm{2}} }{x}+\frac{\left({b}−\mathrm{1}\right)^{\mathrm{2}} }{{b}\left(\mathrm{1}+{b}\right)} \\ $$$$\:\:\:\frac{\left({b}−\mathrm{1}\right)^{\mathrm{2}} }{{b}\left(\mathrm{1}+{b}\right)}\left[\frac{−{x}}{{b}+\mathrm{1}}+\mathrm{1}\right] \\ $$$${forme}\:{a}^{''} {x}+{b}^{''} \:\:\:\left({b}^{''} >\mathrm{0}\:\:\:\:{a}^{''} <\mathrm{0}\right) \\ $$$$\Rightarrow\:\:\:\:{T}^{'} '\:{ander}\:\Gamma^{'} \\ $$
Commented by a.lgnaoui last updated on 31/Mar/23
Commented by alcohol last updated on 31/Mar/23
$${thank}\:{you}\:{very}\:{much} \\ $$