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Question Number 182004 by CrispyXYZ last updated on 03/Dec/22

f(x)=2^x +3^x −6^x Find f(x)_(max)

$${f}\left({x}\right)=\mathrm{2}^{{x}} +\mathrm{3}^{{x}} −\mathrm{6}^{{x}} \\ $$$$\mathrm{Find}\:{f}\left({x}\right)_{\mathrm{max}} \\ $$

Answered by ARUNG_Brandon_MBU last updated on 03/Dec/22

f(x)=2^x +3^x −6^x f ′(x)=2^x ln2+3^x ln3−6^x ln6 f ′(x)∼_(x→x_(max) ) 0 ⇒2^x ln2+3^x ln3−2^x .3^x (ln2+ln3)=0 ⇒ x_(max) =0 ∧ f(x)_(max) =1

$${f}\left({x}\right)=\mathrm{2}^{{x}} +\mathrm{3}^{{x}} −\mathrm{6}^{{x}} \\ $$$${f}\:'\left({x}\right)=\mathrm{2}^{{x}} \mathrm{ln2}+\mathrm{3}^{{x}} \mathrm{ln3}−\mathrm{6}^{{x}} \mathrm{ln6} \\ $$$${f}\:'\left({x}\right)\underset{{x}\rightarrow{x}_{\mathrm{max}} } {\sim}\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}^{{x}} \mathrm{ln2}+\mathrm{3}^{{x}} \mathrm{ln3}−\mathrm{2}^{{x}} .\mathrm{3}^{{x}} \left(\mathrm{ln2}+\mathrm{ln3}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{x}_{\mathrm{max}} =\mathrm{0}\:\wedge\:{f}\left({x}\right)_{\mathrm{max}} =\mathrm{1} \\ $$

Answered by mr W last updated on 03/Dec/22

f(x)=2^x +3^x −2^x 3^x =1−(1−2^x )(1−3^x )≤1−0=1 ⇒f(x)_(max) =1 at 1−2^x =1−3^x =0, i.e. x=0

$${f}\left({x}\right)=\mathrm{2}^{{x}} +\mathrm{3}^{{x}} −\mathrm{2}^{{x}} \mathrm{3}^{{x}} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{1}−\left(\mathrm{1}−\mathrm{2}^{{x}} \right)\left(\mathrm{1}−\mathrm{3}^{{x}} \right)\leqslant\mathrm{1}−\mathrm{0}=\mathrm{1} \\ $$$$\Rightarrow{f}\left({x}\right)_{{max}} =\mathrm{1}\: \\ $$$$\:\:\:\:\:{at}\:\mathrm{1}−\mathrm{2}^{{x}} =\mathrm{1}−\mathrm{3}^{{x}} =\mathrm{0},\:{i}.{e}.\:{x}=\mathrm{0} \\ $$

Answered by SEKRET last updated on 03/Dec/22

2^x =a 3^x =b y=a+b−ab y′(a)=0 y′(b)=0 sistem 1−b=0 1−a=0 M_0 (1 ; 1) y′′(a)=0 y′′(b)=0 y′′(ab)= −1 △= 0 −1= −1<0 2^x =1 3^x =1 x=0 y(0)= 1+1−1=1

$$\:\:\mathrm{2}^{\boldsymbol{\mathrm{x}}} =\boldsymbol{\mathrm{a}}\:\:\:\:\:\mathrm{3}^{\boldsymbol{\mathrm{x}}} =\boldsymbol{\mathrm{b}} \\ $$$$\:\:\:\boldsymbol{\mathrm{y}}=\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{ab}} \\ $$$$\:\:\:\boldsymbol{\mathrm{y}}'\left(\boldsymbol{\mathrm{a}}\right)=\mathrm{0}\:\:\:\:\:\boldsymbol{\mathrm{y}}'\left(\boldsymbol{\mathrm{b}}\right)=\mathrm{0}\:\:\:\boldsymbol{\mathrm{sistem}} \\ $$$$\:\:\:\:\:\:\:\mathrm{1}−\boldsymbol{\mathrm{b}}=\mathrm{0}\:\:\:\:\:\:\:\mathrm{1}−\boldsymbol{\mathrm{a}}=\mathrm{0} \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{M}}_{\mathrm{0}} \left(\mathrm{1}\:;\:\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\mathrm{y}}''\left(\boldsymbol{\mathrm{a}}\right)=\mathrm{0}\:\:\:\:\boldsymbol{\mathrm{y}}''\left(\boldsymbol{\mathrm{b}}\right)=\mathrm{0}\:\:\:\boldsymbol{\mathrm{y}}''\left(\boldsymbol{\mathrm{ab}}\right)=\:−\mathrm{1} \\ $$$$\:\:\bigtriangleup=\:\mathrm{0}\:−\mathrm{1}=\:−\mathrm{1}<\mathrm{0} \\ $$$$\:\:\:\:\:\:\mathrm{2}^{\boldsymbol{\mathrm{x}}} =\mathrm{1}\:\:\:\:\:\:\mathrm{3}^{\boldsymbol{\mathrm{x}}} =\mathrm{1}\:\:\:\:\:\boldsymbol{\mathrm{x}}=\mathrm{0} \\ $$$$\:\:\:\:\boldsymbol{\mathrm{y}}\left(\mathrm{0}\right)=\:\mathrm{1}+\mathrm{1}−\mathrm{1}=\mathrm{1} \\ $$

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