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


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Question Number 190260 by alcohol last updated on 30/Mar/23

$f : [1, 3] \to R, f (x) = \frac{1}{x}$ $A (1, 1)$ $B (1, \frac{1}{3})$ $B^{'} (b, \frac{1}{b}), b ⩾ 1$ $F i n d$ $i . e q u a t i o n o f l i n e {A B}^{'}$ $i i . e q u a t i o n o f t a n g e n t T^{'} t o C_{f} a t p o i n t$ $w i t h x = \frac{1 + b}{2}$ $i i i . S t u d y r e l a t i v e p o s i t i o n s o f L_{A B^{'}}, T^{'} t o C_{f}$
	Answered by a.lgnaoui last updated on 31/Mar/23
	$i.equation of line AB^′ AB^′ :portion of droite AB^′ (y_1 =a^′ x+b^′ ) verifie { ((a^′ +b^′ =1)),((ba^′ +b′=(1/b))) :} y_1 =((−x)/b)+(1/b)+1 ii)equation of tengente T^: to C_f at point:x_0 =((1+b)/2) (df_((x)) /dx)=((−1)/x^2 ) x_0 =((1+b)/2) ⇒ f^′ (x)′=((−4)/((1+b)^2 )) ((f(x)−f(x_0 ))/(x−x_0 ))=((−4)/((1+b)^2 )) y=((4(((1+b)/2)−x))/((1+b)^2 ))+(2/(1+b))=((2(1+b−2x+1+b))/((1+b)^2 )) y_2 =((−4)/((1+b)^2 ))x+(4/(1+b)) iii)Γ^′ is betwen tengente AB′ and C_f ((1/b)>(1/((1+b)^2 ))) with b>1 y_1 −y_2 =(((−x)/b)+((b+1)/b))+((4/((1+b)^2 ))x−(4/(1+b))) =((4/((1+b)^2 ))−(1/b))x+(((b+1)^2 −4b)/(b(1+b))) =((−(b−1)^2 )/(b(1+b)^2 ))x+(((b−1)^2 )/(b(1+b))) (((b−1)^2 )/(b(1+b)))[((−x)/(b+1))+1] forme a^(′′) x+b^(′′) (b^(′′) >0 a^(′′) <0) ⇒ T^′ ′ ander Γ^′$
	$i . e q u a t i o n o f l i n e {A B}^{^{'}}$ ${A B}^{^{'}} : p o r t i o n o f d r o i t e {A B}^{^{'}} (y_{1} = a^{^{'}} x + b^{^{'}})$ $v e r i f i e {\begin{cases} a^{^{'}} + b^{^{'}} = 1 \\ {b a}^{^{'}} + b^{'} = \frac{1}{b} \end{cases}$ $y_{1} = \frac{- x}{b} + \frac{1}{b} + 1$ $i i) e q u a t i o n o f t e n g e n t e T^{:} t o C_{f} a t$ $p o i n t : x_{0} = \frac{1 + b}{2}$ $\frac{{d f}_{(x)}}{d x} = \frac{- 1}{x^{2}} x_{0} = \frac{1 + b}{2} \Rightarrow f^{^{'}} {(x)}^{'} = \frac{- 4}{{(1 + b)}^{2}}$ $\frac{f (x) - f (x_{0})}{x - x_{0}} = \frac{- 4}{{(1 + b)}^{2}}$ $y = \frac{4 (\frac{1 + b}{2} - x)}{{(1 + b)}^{2}} + \frac{2}{1 + b} = \frac{2 (1 + b - 2 x + 1 + b)}{{(1 + b)}^{2}}$ $y_{2} = \frac{- 4}{{(1 + b)}^{2}} x + \frac{4}{1 + b}$ $i i i) Γ^{^{'}} i s b e t w e n t e n g e n t e {A B}^{'} a n d C_{f}$ $(\frac{1}{b} > \frac{1}{{(1 + b)}^{2}}) w i t h b > 1$ $y_{1} - y_{2} = (\frac{- x}{b} + \frac{b + 1}{b}) + (\frac{4}{{(1 + b)}^{2}} x - \frac{4}{1 + b})$ $= (\frac{4}{{(1 + b)}^{2}} - \frac{1}{b}) x + \frac{{(b + 1)}^{2} - 4 b}{b (1 + b)}$ $= \frac{- {(b - 1)}^{2}}{b {(1 + b)}^{2}} x + \frac{{(b - 1)}^{2}}{b (1 + b)}$ $\frac{{(b - 1)}^{2}}{b (1 + b)} [\frac{- x}{b + 1} + 1]$ $f o r m e a^{^{″}} x + b^{^{″}} (b^{^{″}} > 0 a^{^{″}} < 0)$ $Prime causes double exponent: use braces to clarify$
	Commented by a.lgnaoui last updated on 31/Mar/23

	Commented by alcohol last updated on 31/Mar/23

	$t h a n k y o u v e r y m u c h$
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