What is the extreme f(x)=x^3e^{-4x},0<= x<= 2 ?

The extreme f(x)=x^3e^{-4x},0<= x<= 2 is Minimum(0,0),Maximum(3/4 ,(27)/(64e^3)),Minimum(2, 8/(e^8))

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extreme f(x)=x^3e^{-4x},0<= x<= 2

extreme points

f (x) = x^{3} e^{- 4 x}, 0 \leq x \leq 2

Minimum (0, 0), Maximum (\frac{3}{4}, \frac{27}{64 e ^{3}}), Minimum (2, \frac{8}{e ^{8}})

Solution steps

Find the critical points:

x = 0, x = \frac{3}{4}, x = 2

Domain of

x^{3} e^{- 4 x}, 0 \leq x \leq 2 : 0 \leq x \leq 2

Find intervals:

Increasing

: 0 < x < \frac{3}{4},

Decreasing

: \frac{3}{4} < x < 2

Plug the extreme point

x = 0

into

x^{3} e^{- 4 x} \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = \frac{3}{4}

into

x^{3} e^{- 4 x} \Rightarrow y = \frac{27}{64 e ^{3}}

Maximum (\frac{3}{4}, \frac{27}{64 e ^{3}})

Plug the extreme point

x = 2

into

x^{3} e^{- 4 x} \Rightarrow y = \frac{8}{e ^{8}}

Minimum (2, \frac{8}{e ^{8}})

Minimum (0, 0), Maximum (\frac{3}{4}, \frac{27}{64 e ^{3}}), Minimum (2, \frac{8}{e ^{8}})

f (x) = 3 x^{2} + 12 x y + 9 x^{2} + y^{3}

V (u, w) = \frac{1}{2} (3 u + w)

f (x) = (4 - x^{2})^{2}

3 x + 2 y

7 t e^{- \frac{t}{15}}

What is the extreme f(x)=x^3e^{-4x},0<= x<= 2 ?
The extreme f(x)=x^3e^{-4x},0<= x<= 2 is Minimum(0,0),Maximum(3/4 ,(27)/(64e^3)),Minimum(2, 8/(e^8))