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Let 'g' be the function defined by g(x) = (x2-x+1)ex. What is the absolute maximum value of 'g' on the interval [-4,1]?

 Dec 15, 2020
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Let 'g' be the function defined by g(x) = (x2-x+1)\(\cdot e^x\). What is the absolute maximum value of 'g' on the interval [-4,1]?

 

Hello blackobama!

 

\({\color{BrickRed}g_{max_1}=(x^2-x+1)\cdot e^x}=(1^2-1+1)\cdot e^1\)

\(g_{max_1}=e=2.718\)

 

\(\color{BrickRed}g(x)=(x^2-x+1)\cdot e^x\\ g'(x)=u'v+uv'\\g'(x)=(2x-1)e^x+(x^2-x+1)e^x=0\\ x^2+x=0\\\color{blue} x_1=0\\ \color{blue}x_2=-1\)

 

\(g_{max_2}=(x^2-x+1)\cdot e^x=((-1)^2+1+1)\cdot e^{-1}\)

\(g_{max_2}=1.1036\)

 

The absolute highest values

 of \(g(x)=(x^2-x+1)\cdot e^x\)' on the interval [-4,1]

are \(P_{max_1}(-1,\ 1.036]\) and \(P_{max_2}(1,\ e)\)

laugh  !

 Dec 16, 2020
edited by asinus  Dec 16, 2020
edited by asinus  Dec 16, 2020
edited by asinus  Dec 16, 2020
edited by asinus  Dec 16, 2020

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