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# Let 'g' be the function defined by g(x) = (x^2-x+1)e^x. What is the absolute maximum value of 'g' on the interval [-4,1]?

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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 $$P_{max_2}(1,\ e)$$

!

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