What is the extreme f(x)=e^{-x}*(x^2-x+1) ?

The extreme f(x)=e^{-x}*(x^2-x+1) is Minimum(1, 1/e),Maximum(2, 3/(e^2))

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extreme f(x)=e^{-x}*(x^2-x+1)

extreme points

f (x) = e^{- x} \cdot (x^{2} - x + 1)

Minimum (1, \frac{1}{e}), Maximum (2, \frac{3}{e ^{2}})

Solution steps

f^{'} (x) = - e^{- x} (x - 1) (x - 2)

Find intervals:

Decreasing

: - \infty < x < 1,

Increasing

: 1 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = 1

into

e^{- x} \cdot (x^{2} - x + 1) : \frac{1}{e}

Minimum (1, \frac{1}{e})

Plug

x = 2

into

e^{- x} \cdot (x^{2} - x + 1) : \frac{3}{e ^{2}}

Maximum (2, \frac{3}{e ^{2}})

Minimum (1, \frac{1}{e}), Maximum (2, \frac{3}{e ^{2}})

f (x) = - 2 x^{2} + 60 x - 70

y = \frac{cos ( x )}{1 + sin ^{2} ( x )}

f (x, y) = 3 x^{4} + 3 x^{2} y - y^{3}

x^{2} + y^{2} - 2 x - 2 y + 3

f (t) = 5 u t (t - 2)

What is the extreme f(x)=e^{-x}*(x^2-x+1) ?
The extreme f(x)=e^{-x}*(x^2-x+1) is Minimum(1, 1/e),Maximum(2, 3/(e^2))