What is the extreme f(x)=xe^{-3x},0<= x<= 2 ?

The extreme f(x)=xe^{-3x},0<= x<= 2 is Minimum(0,0),Maximum(1/3 , 1/(3e)),Minimum(2, 2/(e^6))

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extreme f(x)=xe^{-3x},0<= x<= 2

extreme points

f (x) = x e^{- 3 x}, 0 \leq x \leq 2

Minimum (0, 0), Maximum (\frac{1}{3}, \frac{1}{3 e}), Minimum (2, \frac{2}{e ^{6}})

Solution steps

Find the critical points:

x = 0, x = \frac{1}{3}, x = 2

Domain of

x e^{- 3 x}, 0 \leq x \leq 2 : 0 \leq x \leq 2

Find intervals:

Increasing

: 0 < x < \frac{1}{3},

Decreasing

: \frac{1}{3} < x < 2

Plug the extreme point

x = 0

into

x e^{- 3 x} \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = \frac{1}{3}

into

x e^{- 3 x} \Rightarrow y = \frac{1}{3 e}

Maximum (\frac{1}{3}, \frac{1}{3 e})

Plug the extreme point

x = 2

into

x e^{- 3 x} \Rightarrow y = \frac{2}{e ^{6}}

Minimum (2, \frac{2}{e ^{6}})

Minimum (0, 0), Maximum (\frac{1}{3}, \frac{1}{3 e}), Minimum (2, \frac{2}{e ^{6}})

f (x) = \frac{x}{x ^{2} + 1} [- 1.2]

\frac{( x ^{2} - 1 )}{( x - 3 ) 6}

f (x) = v (x) = x (7 - x) (15 - x), 0 < x < 7

f (x) = 6 x^{2} - 8 x + 3

f (x, y) = - 25 x^{2} - 25 y^{2} + 300 x + 350 y + 190

What is the extreme f(x)=xe^{-3x},0<= x<= 2 ?
The extreme f(x)=xe^{-3x},0<= x<= 2 is Minimum(0,0),Maximum(1/3 , 1/(3e)),Minimum(2, 2/(e^6))