What are the extrema of $f(x) = e^x(x^2+2x+1)$ ?

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Answer 1 · 2016-04-05T21:18:21

x=-3 or x = -1

$f=e^x, g=x^2+2x+1$

$f'=e^x, g'=2x+2$

$f'(x)=fg'+gf'=e^x (2x+2)+e^x (x^2+2x+1)=0$

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

$e^x (x^2+4x+3)=0$

$e^x(x+3)(x+1)=0$

$e^x = 0 or x+3=0 or x+1 =0$

not possible, $x=-3 or x = -1$

$f(-3)=e^-3(9-6+1)=0.199$ ->max

$f(-1)=e^-1(1-2+1) = 0$ ->min

What are the extrema of f(x) = e^x(x^2+2x+1)f(x) = e^x(x^2+2x+1)?