What is the extreme f(x)=(x^2)/(e^x) ?

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

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extreme f(x)=(x^2)/(e^x)

extreme points

f (x) = \frac{x ^{2}}{e ^{x}}

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

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = 0

into

\frac{x ^{2}}{e ^{x}} : 0

Minimum (0, 0)

Plug

x = 2

into

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

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

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

F (I, J) = (- 3 I + 2 J)

f (x) = x_{1} x

0.5 x^{2} - 130 x + 17555

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

f (x) = x^{4} - 6 x^{2} + 9, 0 \leq x \leq 2

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