What is the extreme f(x)=6x^2-4x+1,0<x<8 ?

The extreme f(x)=6x^2-4x+1,0<x<8 is Minimum(1/3 , 1/3)

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extreme f(x)=6x^2-4x+1,0<x<8

extreme points

f (x) = 6 x^{2} - 4 x + 1, 0 < x < 8

Minimum (\frac{1}{3}, \frac{1}{3})

Solution steps

Find the critical points:

x = \frac{1}{3}

Domain of

6 x^{2} - 4 x + 1, 0 < x < 8 : 0 < x < 8

Find intervals:

Decreasing

: 0 < x < \frac{1}{3},

Increasing

: \frac{1}{3} < x < 8

Plug the extreme point

x = \frac{1}{3}

into

6 x^{2} - 4 x + 1 \Rightarrow y = \frac{1}{3}

Minimum (\frac{1}{3}, \frac{1}{3})

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

f (x) = \frac{69120}{x} + \frac{69120}{y} + 5 x y

f (x) = x^{\frac{2}{3}} (x + 1)

f (x) = x^{2} + 4 x (\frac{4000}{x ^{2}})

f (x) = (1)^{\frac{1}{3}} \cdot (- 9)

What is the extreme f(x)=6x^2-4x+1,0<x<8 ?
The extreme f(x)=6x^2-4x+1,0<x<8 is Minimum(1/3 , 1/3)