What is the extreme f(x)=3|x-1|,0<x<3 ?

The extreme f(x)=3|x-1|,0<x<3 is Minimum(1,0)

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extreme f(x)=3|x-1|,0<x<3

extreme points

f (x) = 3 ∣ x - 1 ∣, 0 < x < 3

Minimum (1, 0)

Solution steps

Find the critical points:

x = 1

Find intervals:

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < 3

Plug the extreme point

x = 1

into

3 ∣ x - 1 ∣ \Rightarrow y = 0

Minimum (1, 0)

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

f (x, y) = x^{3} + y^{3} + x y

f (x) = 2 + x^{\frac{2}{5}}

x^{2} - 4 x + 1

f (x) = x \frac{4}{3} (x - 4)

What is the extreme f(x)=3|x-1|,0<x<3 ?
The extreme f(x)=3|x-1|,0<x<3 is Minimum(1,0)