What is the extreme f(x)=x^3-3x^2-9x+1,0<= x<= 5 ?

The extreme f(x)=x^3-3x^2-9x+1,0<= x<= 5 is Maximum(0,1),Minimum(3,-26),Maximum(5,6)

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extreme f(x)=x^3-3x^2-9x+1,0<= x<= 5

extreme points

f (x) = x^{3} - 3 x^{2} - 9 x + 1, 0 \leq x \leq 5

Maximum (0, 1), Minimum (3, - 26), Maximum (5, 6)

Solution steps

Find the critical points:

x = 0, x = 3, x = 5

Domain of

x^{3} - 3 x^{2} - 9 x + 1, 0 \leq x \leq 5 : 0 \leq x \leq 5

Find intervals:

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < 5

Plug the extreme point

x = 0

into

x^{3} - 3 x^{2} - 9 x + 1 \Rightarrow y = 1

Maximum (0, 1)

Plug the extreme point

x = 3

into

x^{3} - 3 x^{2} - 9 x + 1 \Rightarrow y = - 26

Minimum (3, - 26)

Plug the extreme point

x = 5

into

x^{3} - 3 x^{2} - 9 x + 1 \Rightarrow y = 6

Maximum (5, 6)

Maximum (0, 1), Minimum (3, - 26), Maximum (5, 6)

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

f (x) = x + \frac{4}{x ^{2}}

f (x) = (x - 2) (x + 1) (x + 4)

f (x) = x^{3} - 3 x - 3

x^{2} + y^{2} - x y

What is the extreme f(x)=x^3-3x^2-9x+1,0<= x<= 5 ?
The extreme f(x)=x^3-3x^2-9x+1,0<= x<= 5 is Maximum(0,1),Minimum(3,-26),Maximum(5,6)