What is the extreme f(x)=xe^{-7x},0<= x<= 2 ?

The extreme f(x)=xe^{-7x},0<= x<= 2 is Minimum(0,0),Maximum(1/7 , 1/(7e)),Minimum(2, 2/(e^{14)})

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extreme f(x)=xe^{-7x},0<= x<= 2

extreme points

f (x) = x e^{- 7 x}, 0 \leq x \leq 2

Minimum (0, 0), Maximum (\frac{1}{7}, \frac{1}{7 e}), Minimum (2, \frac{2}{e ^{14}})

Solution steps

Find the critical points:

x = 0, x = \frac{1}{7}, x = 2

Domain of

x e^{- 7 x}, 0 \leq x \leq 2 : 0 \leq x \leq 2

Find intervals:

Increasing

: 0 < x < \frac{1}{7},

Decreasing

: \frac{1}{7} < x < 2

Plug the extreme point

x = 0

into

x e^{- 7 x} \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = \frac{1}{7}

into

x e^{- 7 x} \Rightarrow y = \frac{1}{7 e}

Maximum (\frac{1}{7}, \frac{1}{7 e})

Plug the extreme point

x = 2

into

x e^{- 7 x} \Rightarrow y = \frac{2}{e ^{14}}

Minimum (2, \frac{2}{e ^{14}})

Minimum (0, 0), Maximum (\frac{1}{7}, \frac{1}{7 e}), Minimum (2, \frac{2}{e ^{14}})

f (t) = 2 + 3 o (t)

f (x) = x^{3} - 9 x + 15 x + 3

(x^{2} - 5 x) (y^{2} - 8 y)

T (x, y) = 10 - 2 x^{2} - 3 y^{2}

f (x) = e^{4 x^{2} + 2 y^{2} - 32 x}

What is the extreme f(x)=xe^{-7x},0<= x<= 2 ?
The extreme f(x)=xe^{-7x},0<= x<= 2 is Minimum(0,0),Maximum(1/7 , 1/(7e)),Minimum(2, 2/(e^{14)})