What is the extreme 2x+|x^2-4x+3| ?

The extreme 2x+|x^2-4x+3| is Minimum(1,2)

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extreme 2x+|x^2-4x+3|

extreme points

2 x + x^{2} - 4 x + 3

Minimum (1, 2)

Solution steps

Find the critical points:

x = 1, x = 3

Find intervals:

Decreasing

: - \infty < x < 1,

Increasing

: 1 < x < 3,

Increasing

: 3 < x < \infty

Plug the extreme point

x = 1

into

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

Minimum (1, 2)

f (t) = 2 t - 3 x^{2}

C (x) = 0.6 x^{2} - 96 x + 13223

f (x) = 2 x^{3} - 3 x^{2} - 6

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

y = \frac{2 x ^{2}}{x - 6} [- 3]