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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]?

 Nov 13, 2021
edited by Guest  Nov 13, 2021
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\(g(x)=(x^2-x+1)e^x\)

 

The function is continuous in the interval [-4, 1] so, to find its maximum value, we only have to examine the endpoints of the interval and the points at which the slope (derivative) is 0.

 

First let's find all the values of  x  that make the derivative equal  0 .

 

\(g'(x)=(x^2-x+1)(e^x)+(e^x)(2x-1)\)

 

Set the derivative equal to 0 and solve for  x :

 

\((x^2-x+1)(e^x)+(e^x)(2x-1)=0\\~\\ e^x(\ (x^2-x+1)+(2x-1)\ )=0\\~\\ e^x(\ x^2+x\ )=0\\~\\ e^x(x)(x+1)=0\)

 

Set each factor equal to zero and solve each equation for  x:

 

\(\begin{array}{} e^x=0&\qquad\text{or}\qquad &x=0&\qquad\text{or}\qquad& x+1=0\\~\\ \text{no solution}&&x=0&&x=-1 \end{array}\)

 

So only the following x values have the possibility to produce the maximum  g  in the interval:  -4, -1, 0 1

 

\(\begin{array}{} g(-4)&=& ((-4)^2-(-4)+1)e^{-4}&=& \frac{21}{e^4}& \approx&0.385\\~\\ g(-1)& =& ((-1)^2-(-1)+1)e^{-1}& =&\frac{3}{e^1}&\approx& 1.104\\~\\ g(0)&=& ((0)^2-(0)+1)e^{0}& =& (1)1& =& 1\\~\\ g(1)&=& ((1)^2-(1)+1)e^{1}& =& (1)e^1& \approx& 2.718 \end{array}\)

 

The maximum value of  g  in the interval  [-4, 1]  is  2.718  and it occurs when  x = 1

 

Check: https://www.desmos.com/calculator/eesnhly6qa

 Nov 13, 2021

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