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Question Number 196710 by tri26112004 last updated on 30/Aug/23

Give the function:  f(x)=(x−1)(x−2)^2 (x−3)^3 (x−4)^4 ...(x−2022)^(2022)   Find extremes of f(x)¿

$${Give}\:{the}\:{function}: \\ $$$${f}\left({x}\right)=\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)^{\mathrm{2}} \left({x}−\mathrm{3}\right)^{\mathrm{3}} \left({x}−\mathrm{4}\right)^{\mathrm{4}} ...\left({x}−\mathrm{2022}\right)^{\mathrm{2022}} \\ $$$${Find}\:{extremes}\:{of}\:{f}\left({x}\right)¿ \\ $$

Commented by mr W last updated on 30/Aug/23

i don′t think they can be exactly   determined, because we can′t exactly  solve the equation f′(x)=0, i.e.  ln ∣x−1∣+2ln ∣x−2∣+3ln ∣x−3∣+...+2022ln ∣x−2022∣=0.

$${i}\:{don}'{t}\:{think}\:{they}\:{can}\:{be}\:{exactly}\: \\ $$$${determined},\:{because}\:{we}\:{can}'{t}\:{exactly} \\ $$$${solve}\:{the}\:{equation}\:{f}'\left({x}\right)=\mathrm{0},\:{i}.{e}. \\ $$$$\mathrm{ln}\:\mid{x}−\mathrm{1}\mid+\mathrm{2ln}\:\mid{x}−\mathrm{2}\mid+\mathrm{3ln}\:\mid{x}−\mathrm{3}\mid+...+\mathrm{2022ln}\:\mid{x}−\mathrm{2022}\mid=\mathrm{0}. \\ $$

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