What is the extreme f(x)=(x^3)/3-x^2-3x-1,2<= x<= 7 ?

The extreme f(x)=(x^3)/3-x^2-3x-1,2<= x<= 7 is Maximum(2,-25/3),Minimum(3,-10),Maximum(7, 130/3)

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extreme f(x)=(x^3)/3-x^2-3x-1,2<= x<= 7

extreme points

f (x) = \frac{x ^{3}}{3} - x^{2} - 3 x - 1, 2 \leq x \leq 7

Maximum (2, - \frac{25}{3}), Minimum (3, - 10), Maximum (7, \frac{130}{3})

Solution steps

Find the critical points:

x = 2, x = 3, x = 7

Domain of

\frac{x ^{3}}{3} - x^{2} - 3 x - 1, 2 \leq x \leq 7 : 2 \leq x \leq 7

Find intervals:

Decreasing

: 2 < x < 3,

Increasing

: 3 < x < 7

Plug the extreme point

x = 2

into

\frac{x ^{3}}{3} - x^{2} - 3 x - 1 \Rightarrow y = - \frac{25}{3}

Maximum (2, - \frac{25}{3})

Plug the extreme point

x = 3

into

\frac{x ^{3}}{3} - x^{2} - 3 x - 1 \Rightarrow y = - 10

Minimum (3, - 10)

Plug the extreme point

x = 7

into

\frac{x ^{3}}{3} - x^{2} - 3 x - 1 \Rightarrow y = \frac{130}{3}

Maximum (7, \frac{130}{3})

Maximum (2, - \frac{25}{3}), Minimum (3, - 10), Maximum (7, \frac{130}{3})

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

f (x) = 416 - x x

f (x) = x^{\frac{4}{5}} - 4

f (x) = (\frac{5 x ^{4}}{ln ( x )})

f (x) = - (\frac{4}{x ^{2}}), (- 3, - 1)

What is the extreme f(x)=(x^3)/3-x^2-3x-1,2<= x<= 7 ?
The extreme f(x)=(x^3)/3-x^2-3x-1,2<= x<= 7 is Maximum(2,-25/3),Minimum(3,-10),Maximum(7, 130/3)