# What are the critical points of f(x) = sqrt((x^2 - 4x +5))?

Apr 16, 2018

The critical point of$f \left(x\right)$ is $\left(2 , 1\right)$

#### Explanation:

$y = \sqrt{{x}^{2} - 4 x + 5}$

Differentiate

$y ' = \frac{2 x - 4}{2 \sqrt{{x}^{2} - 4 x + 5}}$

now critical points are points at which $y ' = 0$ or the tangent is vertical at that point i.e. $y ' = \frac{1}{0}$

$y ' = 0$$\textcolor{b l u e}{\rightarrow}$$2 x - 4 = 0$

$x = 2$

$y ' = \frac{1}{0}$$\textcolor{b l u e}{\rightarrow}$${x}^{2} - 4 x + 5 = 0$

$x = 2 \pm i$ which it refused as it's not a real number

So there is one critical point at $x = 2$
$f \left(2\right) = 1$

The critical point of$f \left(x\right)$ is $\left(2 , 1\right)$