Find the relative maximum of f(x)=1+(2+x)e^-x

Question

Find the relative maximum.

Answer 1

f(x) = 1 + (2+x)e^-x

= 1 + 2e^-x + xe^-x

take the derivative and set = 0

f'(x) = -2e^-x -xe^-x + e^-x=0

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

= e^-x(-x-1) = 0

-x-1 = 0

x= -1

f(-1) = 1 + (2-1)e^--1 = 1 +e = about 1+2.718 = about 3.718

Answer 2

f'(x) = 0

0+d/dx(2+x)*e^-x+(2+x)*d/dx(e^-x)=0

1*e^-x+(2+x)(-e^-x)=0

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

e^-x(-1-x)=0

x = -1

Relative maximum = f(-1) = 1+(2-1)e^1 = 1+ e = 3.718

Hope this helps!

Answer 3

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