What is the extreme y=|x^2-3x| ?

The extreme y=|x^2-3x| is Minimum(0,0),Maximum(3/2 , 9/4),Minimum(3,0)

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extreme y=|x^2-3x|

extreme points

y = x^{2} - 3 x

Minimum (0, 0), Maximum (\frac{3}{2}, \frac{9}{4}), Minimum (3, 0)

Solution steps

Find the critical points:

x = 0, x = \frac{3}{2}, x = 3

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

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

Decreasing

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

Increasing

: 3 < x < \infty

Plug the extreme point

x = 0

into

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

Minimum (0, 0)

Plug the extreme point

x = \frac{3}{2}

into

x^{2} - 3 x \Rightarrow y = \frac{9}{4}

Maximum (\frac{3}{2}, \frac{9}{4})

Plug the extreme point

x = 3

into

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

Minimum (3, 0)

Minimum (0, 0), Maximum (\frac{3}{2}, \frac{9}{4}), Minimum (3, 0)

y = 9 - 6 x^{2}, - 2 \leq x \leq 3

f (x) = 2 x^{3} + 4 x^{2} - 1

49 x + \frac{1}{x}

f (x) = - 0.1 x^{2} + 1.2 x + 5

7 (x - 8)^{3}

What is the extreme y=|x^2-3x| ?
The extreme y=|x^2-3x| is Minimum(0,0),Maximum(3/2 , 9/4),Minimum(3,0)