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

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Answer 1 · 2018-04-16T15:38:21

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

$y=sqrt(x^2-4x+5)$

Differentiate

$y'=(2x-4)/(2sqrt(x^2-4x+5))$

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

$y'=0$ $color(blue)rarr$ $2x-4=0$

$x=2$

$y'=1/0$ $color(blue)rarr$ $x^2-4x+5=0$

$x=2+-i$ which it refused as it's not a real number

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

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

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