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

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

Answered by ARUNG_Brandon_MBU last updated on 03/Dec/22

$$f\left(x\right)=2^x+3^x-6^x$$$$f{}^{\prime }\left(x\right)=2^x\ln 2+3^x\ln 3-6^x\ln 6$$$$f{}^{\prime }\left(x\right)\underset{x\to x_{\max }}{\sim }0$$$$\Rightarrow 2^x\ln 2+3^x\ln 3-2^x.3^x(\ln 2+\ln 3)=0$$$$\Rightarrow x_{\max }=0\wedge f{\left(x\right)}_{\max }=1$$

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

$$f\left(x\right)=2^x+3^x-2^x{3}^x$$$$=1-(1-2^x)(1-3^x)⩽1-0=1$$$$\Rightarrow f{\left(x\right)}_{max}=1$$$$at1-2^x=1-3^x=0,i.e.x=0$$

Answered by SEKRET last updated on 03/Dec/22

$$2^{\mathbf{x}}=\mathbf{a}{3}^{\mathbf{x}}=\mathbf{b}$$$$\mathbf{y}=\mathbf{a}+\mathbf{b}-\mathbf{ab}$$$${\mathbf{y}}^{\prime }\left(\mathbf{a}\right)=0{\mathbf{y}}^{\prime }\left(\mathbf{b}\right)=0\mathbf{sistem}$$$$1-\mathbf{b}=01-\mathbf{a}=0$$$${\mathbf{M}}_0(1;1)$$$${\mathbf{y}}^{″}\left(\mathbf{a}\right)=0{\mathbf{y}}^{″}\left(\mathbf{b}\right)=0{\mathbf{y}}^{″}\left(\mathbf{ab}\right)=-1$$$$△=0-1=-1<0$$$$2^{\mathbf{x}}=13^{\mathbf{x}}=1\mathbf{x}=0$$$$\mathbf{y}\left(0\right)=1+1-1=1$$

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