What is the extreme f(x)=|x|+x^2-5x+6,(-1,4) ?

The extreme f(x)=|x|+x^2-5x+6,(-1,4) is Minimum(2,2)

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extreme f(x)=|x|+x^2-5x+6,(-1,4)

extreme points

f (x) = ∣ x ∣ + x^{2} - 5 x + 6, (- 1, 4)

Minimum (2, 2)

Solution steps

Find the critical points:

x = 0, x = 2

Find intervals:

Decreasing

: - 1 < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < 4

Plug the extreme point

x = 2

into

∣ x ∣ + x^{2} - 5 x + 6 \Rightarrow y = 2

Minimum (2, 2)

f (x, y) = 7 x^{2} + 6 x y + 5 y^{2} + 4 x y^{2} + 7 x^{2} y

x^{2} + x y + y^{2} + 8 y

5 - x

3 x^{3} - 3 x^{2} - 3 x + 5, - 1 \leq x \leq 2

f (x) = 8 x - 2 x^{2} + 12 x y - 12 y^{2} + y

What is the extreme f(x)=|x|+x^2-5x+6,(-1,4) ?
The extreme f(x)=|x|+x^2-5x+6,(-1,4) is Minimum(2,2)