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Question Number 39338 by rahul 19 last updated on 05/Jul/18

$$\mathrm{If}\mathrm{f}\left(x\right)={\int }_0^4{\mathrm{e}}^{\mid \mathrm{t}-x\mid }\mathrm{dt}(0⩽x⩽4),$$$$\mathrm{maximum}\mathrm{value}\mathrm{of}\mathrm{f}\left(x\right)\mathrm{is}=?$$

Commented by math khazana by abdo last updated on 06/Jul/18

$$wehavef\left(x\right)={\int }_0^x{e}^{x-t\mid }dt+{\int }_x^4{e}^{t-x}dt$$$$=e^x{[-e^{-t}]}_0^x+e^{-x}{\left[e^t\right]}_x^4$$$$=e^x(1-e^{-x})+e^{-x}(e^4-e^x)$$$$=e^x-1+e^4{e}^{-x}-1=e^x+e^4{e}^{-x}-2$$$$f^{{}^{\prime }}\left(x\right)=e^x-e^4{e}^{-x}=e^x(1-e^4{e}^{-2x})so$$$$f^{{}^{\prime }}\left(x\right)=0\iff 1-e^4{e}^{-2x}=0\iff e^{-2x}=e^{-4}\iff x=2$$$$f^{{}^{\prime }}\left(x\right)⩾0\iff 1-e^4{e}^{-2x}⩾0\iff e^4{e}^{-2x}⩽1\iff$$$$e^{-2x}⩽e^{-4}\iff -2x⩽-4\iff x⩾2sofis$$$$increazingon[2,4]decreasingon[0,2]\Rightarrow$$$$maxf\left(x\right)=f\left(0\right)orf\left(4\right)but$$$$f\left(0\right)=e^4-1andf\left(4\right)=e^4-1=f\left(0\right)\Rightarrow$$$$maxf\left(x\right)=f\left(0\right)=e^4-1$$

Commented by rahul 19 last updated on 05/Jul/18

prof ans is (e⁴ - 1 )

Answered by ajfour last updated on 05/Jul/18

$$f\left(x\right){\mid }_{max}={\int }_0^4{e}^{t}dt=e^4-1.$$

Commented by rahul 19 last updated on 05/Jul/18

Thank you sir .

Answered by MJS last updated on 06/Jul/18

$$\frac{d}{dx}[\int {\mathrm{e}}^{\mid t-x\mid }dt]=-{\mathrm{e}}^{\mid t-x\mid }\Rightarrow$$$$\Rightarrow \frac{d}{dx}\left[\underset{0}{\stackrel{4}{\int }}{\mathrm{e}}^{\mid t-x\mid }dt\right]=-{\mathrm{e}}^{\mid 4-x\mid }+{\mathrm{e}}^{\mid x\mid }=f^{\prime }\left(x\right)$$$$f^{″}\left(x\right)=\mathrm{sign}(4-x){\mathrm{e}}^{\mid 4-x\mid }+\mathrm{sign}\left(x\right){\mathrm{e}}^{\mid x\mid }$$$$f^{\prime }\left(x\right)=0\Rightarrow \mid x\mid =\mid 4-x\mid \Rightarrow x=2$$$$f^{″}\left(2\right)={2\mathrm{e}}^2>0\Rightarrow f\left(x\right)\mathrm{has}\mathrm{an}\mathrm{absolute}\mathrm{minimum}$$$$\mathrm{at}x=2\Rightarrow \mathrm{maximum}\mathrm{is}\mathrm{at}\mathrm{the}\mathrm{borders}\mathrm{of}$$$$\mathrm{given}\mathrm{interval}\Rightarrow \max \left(f\left(x\right)\right)=f\left(0\right)=f\left(4\right)={\mathrm{e}}^4-1$$$$$$

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