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

Answered by liberty last updated on 18/Jun/21

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

Answered by TheHoneyCat last updated on 18/Jun/21

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

Commented by abdullahoudou last updated on 18/Jun/21

$$youmaymean12\times 4/3$$$$$$

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

$$\iff yx+y=x^2+12x+20$$$$\Rightarrow x^2+(12-y)x+20-y=0$$$$\mathrm{\Delta }={(12-y)}^2-4(20-y)⩾0$$$$\Rightarrow y^2-20y+64⩾0$$$$\Rightarrow (y-4)(y-16)⩾0$$$$\Rightarrow y⩽4\cup y⩾16$$$$then{y}_{min}=16$$

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