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Question Number 143781 by bemath last updated on 18/Jun/21

Answered by liberty last updated on 18/Jun/21

let x+1= t⇒f(t−1)=y=(((t+9)(t+1))/t) y=t+10+(9/t) .⇒ t+(9/t) has minimum 6 for t=3 so f(t)_(min) =3+10+(9/3)=16

$$\mathrm{let}\:\mathrm{x}+\mathrm{1}=\:\mathrm{t}\Rightarrow\mathrm{f}\left(\mathrm{t}−\mathrm{1}\right)=\mathrm{y}=\frac{\left(\mathrm{t}+\mathrm{9}\right)\left(\mathrm{t}+\mathrm{1}\right)}{\mathrm{t}} \\ $$$$\mathrm{y}=\mathrm{t}+\mathrm{10}+\frac{\mathrm{9}}{\mathrm{t}}\:.\Rightarrow\:\mathrm{t}+\frac{\mathrm{9}}{\mathrm{t}}\:\mathrm{has}\:\mathrm{minimum}\:\mathrm{6}\:\mathrm{for}\:\mathrm{t}=\mathrm{3}\: \\ $$$$\mathrm{so}\:\mathrm{f}\left(\mathrm{t}\right)_{\mathrm{min}} =\mathrm{3}+\mathrm{10}+\frac{\mathrm{9}}{\mathrm{3}}=\mathrm{16} \\ $$

Answered by TheHoneyCat last updated on 18/Jun/21

(df/dx)(x)=(((2x+12)(x+1)−(x^2 +12x+20))/((x+1)^2 )) =((x^2 +2x−8)/((x+1)^2 )) =(((x−2)(x+4))/((x+1)^2 )) ∀x∈R_+ ((x+4)/((x+1)^2 ))>0 therefore: x∈[0,2[ ⇒ (df/dx)(x)<0 x∈]2,+∞[ ⇒ (df/dx)(x)>0 and (df/dx)(2)=0 so f′s minimal value on R_+ is reached for x=2 Min_R_+ f=f(2)=((12×4)/2)=24_■

$$\frac{{df}}{{dx}}\left({x}\right)=\frac{\left(\mathrm{2}{x}+\mathrm{12}\right)\left({x}+\mathrm{1}\right)−\left({x}^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{20}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{8}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{4}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\forall{x}\in\mathbb{R}_{+} \frac{{x}+\mathrm{4}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }>\mathrm{0} \\ $$$$ \\ $$$$\mathrm{therefore}: \\ $$$${x}\in\left[\mathrm{0},\mathrm{2}\left[\:\Rightarrow\:\frac{{df}}{{dx}}\left({x}\right)<\mathrm{0}\right.\right. \\ $$$$\left.{x}\in\right]\mathrm{2},+\infty\left[\:\Rightarrow\:\frac{{df}}{{dx}}\left({x}\right)>\mathrm{0}\right. \\ $$$$\mathrm{and}\:\:\frac{{df}}{{dx}}\left(\mathrm{2}\right)=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{so}\:{f}'\mathrm{s}\:\mathrm{minimal}\:\mathrm{value}\:\mathrm{on}\:\mathbb{R}_{+} \:\mathrm{is}\:\mathrm{reached}\:\mathrm{for}\:{x}=\mathrm{2} \\ $$$$ \\ $$$$\mathrm{Min}_{\mathbb{R}_{+} } {f}={f}\left(\mathrm{2}\right)=\frac{\mathrm{12}×\mathrm{4}}{\mathrm{2}}=\mathrm{24}_{\blacksquare} \\ $$

Commented by abdullahoudou last updated on 18/Jun/21

you may mean 12×4/3

$${you}\:{may}\:{mean}\:\mathrm{12}×\mathrm{4}/\mathrm{3} \\ $$$$ \\ $$

Commented by TheHoneyCat last updated on 19/Jun/21

Of course, I had to make a mistake at the very last line. �� Yes indeed, f(2)=12x4/3=16. Thanks @abdullahoudou, for paying attention to the little details. ��

Answered by john_santu last updated on 19/Jun/21

⇔yx+y=x^2 +12x+20 ⇒x^2 +(12−y)x+20−y=0 Δ=(12−y)^2 −4(20−y)≥0 ⇒y^2 −20y+64≥0 ⇒(y−4)(y−16)≥0 ⇒y≤4 ∪ y≥ 16 then y_(min) = 16

$$\Leftrightarrow{yx}+{y}={x}^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{20} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +\left(\mathrm{12}−{y}\right){x}+\mathrm{20}−{y}=\mathrm{0} \\ $$$$\Delta=\left(\mathrm{12}−{y}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{20}−{y}\right)\geqslant\mathrm{0} \\ $$$$\Rightarrow{y}^{\mathrm{2}} −\mathrm{20}{y}+\mathrm{64}\geqslant\mathrm{0} \\ $$$$\Rightarrow\left({y}−\mathrm{4}\right)\left({y}−\mathrm{16}\right)\geqslant\mathrm{0} \\ $$$$\Rightarrow{y}\leqslant\mathrm{4}\:\cup\:{y}\geqslant\:\mathrm{16}\: \\ $$$${then}\:{y}_{{min}} \:=\:\mathrm{16}\: \\ $$

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