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

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

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

extreme points

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

Minimum (3, - 26)

Solution steps

Find the critical points:

x = 3

Domain of

x^{3} - 3 x^{2} - 9 x + 1, (0, 5) : 0 < x < 5

Find intervals:

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < 5

Plug the extreme point

x = 3

into

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

Minimum (3, - 26)

f (x) x^{3} - 6 x^{2} + 9 x + 1

f (x, y) = 9 - x^{2} - y^{2} - 4 y^{2}

- 0.02 x^{2} + 1.6 x + 98.2

9 x^{3} - 54 x^{2} + 81 x + 13, - 6 \leq x \leq 2

f (x) = \frac{( x ^{2} - 121 )}{2 x ^{2}}

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