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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

extreme points

f (x) = ∣ x ∣ + x^{2} - 5 x + 6

Minimum (2, 2)

Solution steps

Find the critical points:

x = 0, x = 2

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug the extreme point

x = 2

into

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

Minimum (2, 2)

f (x, y) = 60 x^{\frac{1}{3}} y^{\frac{2}{3}}

x^{2} (x^{2} - 2)

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

- 3 x^{3}

y = x^{5} - 5 x^{4}

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