What is the extreme f(x)=x^3-9x^2+24x-2,0<= x<= 5 ?

The extreme f(x)=x^3-9x^2+24x-2,0<= x<= 5 is Minimum(0,-2),Maximum(2,18),Minimum(4,14),Maximum(5,18)

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extreme f(x)=x^3-9x^2+24x-2,0<= x<= 5

extreme points

f (x) = x^{3} - 9 x^{2} + 24 x - 2, 0 \leq x \leq 5

Minimum (0, - 2), Maximum (2, 18), Minimum (4, 14), Maximum (5, 18)

Solution steps

Find the critical points:

x = 0, x = 2, x = 4, x = 5

Domain of

x^{3} - 9 x^{2} + 24 x - 2, 0 \leq x \leq 5 : 0 \leq x \leq 5

Find intervals:

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < 4,

Increasing

: 4 < x < 5

Plug the extreme point

x = 0

into

x^{3} - 9 x^{2} + 24 x - 2 \Rightarrow y = - 2

Minimum (0, - 2)

Plug the extreme point

x = 2

into

x^{3} - 9 x^{2} + 24 x - 2 \Rightarrow y = 18

Maximum (2, 18)

Plug the extreme point

x = 4

into

x^{3} - 9 x^{2} + 24 x - 2 \Rightarrow y = 14

Minimum (4, 14)

Plug the extreme point

x = 5

into

x^{3} - 9 x^{2} + 24 x - 2 \Rightarrow y = 18

Maximum (5, 18)

Minimum (0, - 2), Maximum (2, 18), Minimum (4, 14), Maximum (5, 18)

f (x) = 9 x^{5} - 2 x^{4} - 2 x^{3} + 4

3 x^{2} - 12 x - 15

f (x) = x y + \frac{9}{x} + \frac{3}{y}

f (x) = x^{3} - 3 x^{2} + b

x^{3} + 6 x^{2} + 12 x + 2

What is the extreme f(x)=x^3-9x^2+24x-2,0<= x<= 5 ?
The extreme f(x)=x^3-9x^2+24x-2,0<= x<= 5 is Minimum(0,-2),Maximum(2,18),Minimum(4,14),Maximum(5,18)