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

The extreme f(x)=x^2e^{1-x} is Minimum(0,0),Maximum(2, 4/e)

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

extreme points

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

Minimum (0, 0), Maximum (2, \frac{4}{e})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = 0

into

x^{2} e^{1 - x} : 0

Minimum (0, 0)

Plug

x = 2

into

x^{2} e^{1 - x} : \frac{4}{e}

Maximum (2, \frac{4}{e})

Minimum (0, 0), Maximum (2, \frac{4}{e})

(3 - x) (x - 5)^{2}

f (x, y) = \frac{1}{x} + \frac{1}{y} - x y + 3

f (x) = x^{3} + y^{3} + 9 x^{2} - 9 y^{2} - 2

f (x) = \frac{( x )}{( 6 x + 2 )}, 5 \leq x \leq 8

f (x, y) = (x - 1)^{3} + (x - 1)^{2} + (y + 1)^{2}

What is the extreme f(x)=x^2e^{1-x} ?
The extreme f(x)=x^2e^{1-x} is Minimum(0,0),Maximum(2, 4/e)