Question-143781

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

let x + 1 = t \Rightarrow f (t - 1) = y = \frac{(t + 9) (t + 1)}{t}

y = t + 10 + \frac{9}{t} . \Rightarrow t + \frac{9}{t} has minimum 6 for t = 3

so f {(t)}_{\min} = 3 + 10 + \frac{9}{3} = 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{d f}{d x} (x) = \frac{(2 x + 12) (x + 1) - (x^{2} + 12 x + 20)}{{(x + 1)}^{2}}

= \frac{x^{2} + 2 x - 8}{{(x + 1)}^{2}}

= \frac{(x - 2) (x + 4)}{{(x + 1)}^{2}}

\forall x \in R_{+} \frac{x + 4}{{(x + 1)}^{2}} > 0

therefore :

x \in [0, 2 [\Rightarrow \frac{d f}{d x} (x) < 0

x \in] 2, + \infty [\Rightarrow \frac{d f}{d x} (x) > 0

and \frac{d f}{d x} (2) = 0

so f^{'} s minimal value on R_{+} is reached for x = 2

{Min}_{R_{+}} f = f (2) = \frac{12 \times 4}{2} = 24_{}

Commented by abdullahoudou last updated on 18/Jun/21

y o u m a y m e a n 12 \times 4 / 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 y x + y = x^{2} + 12 x + 20

\Rightarrow x^{2} + (12 - y) x + 20 - y = 0

Δ = {(12 - y)}^{2} - 4 (20 - y) ⩾ 0

\Rightarrow y^{2} - 20 y + 64 ⩾ 0

\Rightarrow (y - 4) (y - 16) ⩾ 0

\Rightarrow y ⩽ 4 \cup y ⩾ 16

t h e n y_{m i n} = 16

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