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

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

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

extreme points

f (x) = \frac{x ^{3}}{3} + 2 x^{2} - 4, - 2 \leq x \leq 2

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

Solution steps

Find the critical points:

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

Domain of

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

Find intervals:

Decreasing

: - 2 < x < 0,

Increasing

: 0 < x < 2

Plug the extreme point

x = - 2

into

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

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

Plug the extreme point

x = 0

into

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

Minimum (0, - 4)

Plug the extreme point

x = 2

into

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

Maximum (2, \frac{20}{3})

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

f (x) = \frac{1}{4} x^{4} - 2 x^{3} - 18 x^{2} + 216 x

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

f (x) = - 2 x y - 8 x^{2} - 2 y^{2} - 86 x + 8 y + 9

f (x) = 3 x + 2

y = 2 x + sin (4 x), - \frac{π}{3} \leq x \leq \frac{π}{3}

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