What is the extreme f(x)=(x^3)/3-3x^2+8x-2,0<= x<= 3 ?

The extreme f(x)=(x^3)/3-3x^2+8x-2,0<= x<= 3 is Minimum(0,-2),Maximum(2, 14/3),Minimum(3,4)

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extreme f(x)=(x^3)/3-3x^2+8x-2,0<= x<= 3

extreme

Minimum (0, - 2), Maximum (2, \frac{14}{3}), Minimum (3, 4)

Solution steps

Find the critical points:

x = 0, x = 2, x = 3

Domain of

\frac{x ^{3}}{3} - 3 x^{2} + 8 x - 2, 0 \leq x \leq 3 : 0 \leq x \leq 3

Find intervals:

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < 3

Plug the extreme point

x = 0

into

\frac{x ^{3}}{3} - 3 x^{2} + 8 x - 2 \Rightarrow y = - 2

Minimum (0, - 2)

Plug the extreme point

x = 2

into

\frac{x ^{3}}{3} - 3 x^{2} + 8 x - 2 \Rightarrow y = \frac{14}{3}

Maximum (2, \frac{14}{3})

Plug the extreme point

x = 3

into

\frac{x ^{3}}{3} - 3 x^{2} + 8 x - 2 \Rightarrow y = 4

Minimum (3, 4)

Minimum (0, - 2), Maximum (2, \frac{14}{3}), Minimum (3, 4)

What is the extreme f(x)=(x^3)/3-3x^2+8x-2,0<= x<= 3 ?
The extreme f(x)=(x^3)/3-3x^2+8x-2,0<= x<= 3 is Minimum(0,-2),Maximum(2, 14/3),Minimum(3,4)