+0  
 
+1
1927
1
avatar+13 

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
 #1
avatar+11238 
+1

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

58 Online Users

avatar
avatar
avatar
avatar