What is the extreme f(x)=(x^3)/3-2x^2+4,-2<= x<= 1 ?

The extreme f(x)=(x^3)/3-2x^2+4,-2<= x<= 1 is Minimum(-2,-20/3),Maximum(0,4),Minimum(1, 7/3)

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extreme f(x)=(x^3)/3-2x^2+4,-2<= x<= 1

extreme

Minimum (- 2, - \frac{20}{3}), Maximum (0, 4), Minimum (1, \frac{7}{3})

Solution steps

Find the critical points:

x = - 2, x = 0, x = 1

Domain of

\frac{x ^{3}}{3} - 2 x^{2} + 4, - 2 \leq x \leq 1 : - 2 \leq x \leq 1

Find intervals:

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 1

Plug the extreme point

x = - 2

into

\frac{x ^{3}}{3} - 2 x^{2} + 4 \Rightarrow y = - \frac{20}{3}

Minimum (- 2, - \frac{20}{3})

Plug the extreme point

x = 0

into

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

Maximum (0, 4)

Plug the extreme point

x = 1

into

\frac{x ^{3}}{3} - 2 x^{2} + 4 \Rightarrow y = \frac{7}{3}

Minimum (1, \frac{7}{3})

Minimum (- 2, - \frac{20}{3}), Maximum (0, 4), Minimum (1, \frac{7}{3})

What is the extreme f(x)=(x^3)/3-2x^2+4,-2<= x<= 1 ?
The extreme f(x)=(x^3)/3-2x^2+4,-2<= x<= 1 is Minimum(-2,-20/3),Maximum(0,4),Minimum(1, 7/3)