What is the extreme+sqrt(|x^2-3x+2|) ?

The extreme+sqrt(|x^2-3x+2|) is Minimum(1,0),Maximum(3/2 , 1/2),Minimum(2,0)

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extreme+sqrt(|x^2-3x+2|)

extreme points

+ ∣ x^{2} - 3 x + 2 ∣

Minimum (1, 0), Maximum (\frac{3}{2}, \frac{1}{2}), Minimum (2, 0)

Solution steps

Find the critical points:

x = 1, x = \frac{3}{2}, x = 2

Find intervals:

Decreasing

: - \infty < x < 1,

Increasing

: 1 < x < \frac{3}{2},

Decreasing

: \frac{3}{2} < x < 2,

Increasing

: 2 < x < \infty

Plug the extreme point

x = 1

into

+ ∣ x^{2} - 3 x + 2 ∣ \Rightarrow y = 0

Minimum (1, 0)

Plug the extreme point

x = \frac{3}{2}

into

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

Maximum (\frac{3}{2}, \frac{1}{2})

Plug the extreme point

x = 2

into

+ ∣ x^{2} - 3 x + 2 ∣ \Rightarrow y = 0

Minimum (2, 0)

Minimum (1, 0), Maximum (\frac{3}{2}, \frac{1}{2}), Minimum (2, 0)

f (x) = (- 5 x^{2} + 10 x y - 20 x - 7 y^{2} + 240 y - 5300)

f (x, y) = 3 x y + 6 y - 5 x

y = I n ∣ 1 - x ∣

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

f (x) = 36 - t^{2}

What is the extreme+sqrt(|x^2-3x+2|) ?
The extreme+sqrt(|x^2-3x+2|) is Minimum(1,0),Maximum(3/2 , 1/2),Minimum(2,0)