# Determine the local max and/or min and intervals of increase and decrease for the function f(x)=√(x^2 - 2x +2) ?

May 7, 2018

$f$ is decreasing in $\left(- \infty , 1\right]$ and increasing in $\left[1 , + \infty\right)$ so $f$ has a local and global $\min$ at ${x}_{0} = 1$ , $f \left(1\right) = 1$
$\to f \left(x\right) \ge f \left(1\right) = 1 > 0$ , $x$$\in$$\mathbb{R}$

#### Explanation:

$f \left(x\right) = \sqrt{{x}^{2} - 2 x + 2}$ , ${D}_{f} = \mathbb{R}$

$\forall$$x$$\in$$\mathbb{R}$,

f'(x)=((x^2-2x+2)')/(2sqrt(x^2-2x+2) $=$

(2x-2)/(2sqrt(x^2-2x+2) $=$

(x-1)/(sqrt(x^2-2x+2)

with $f ' \left(x\right) = 0 \iff \left(x = 1\right)$

• $x$$\in$$\left(- \infty , 1\right)$, $f ' \left(x\right) < 0$ so $f$ is decreasing in $\left(- \infty , 1\right]$
• $x$$\in$$\left(1 , + \infty\right)$ , $f ' \left(x\right) > 0$ so $f$ is increasing in $\left[1 , + \infty\right)$

$f$ is decreasing in $\left(- \infty , 1\right]$ and increasing in $\left[1 , + \infty\right)$ so $f$ has a local and global $\min$ at ${x}_{0} = 1$ , $f \left(1\right) = 1$
$\to f \left(x\right) \ge f \left(1\right) = 1 > 0$ , $x$$\in$$\mathbb{R}$

Graphical help
graph{sqrt(x^2-2x+2) [-10, 10, -5, 5]}