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

Question

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

Charles Arocha · Accepted Answer

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)#

Christopher Allers · Answer

undefined To find the critical points of $ f(x) = \sqrt{x^2 - 4x + 5} $, first, find its derivative, then solve for where the derivative equals zero or is undefined.

1. Find the derivative of $ f(x) $:
$$ f'(x) = \frac{1}{2\sqrt{x^2 - 4x + 5}} \cdot (2x - 4) $$

2. Set $ f'(x) $ equal to zero and solve for $ x $:
$$ \frac{1}{2\sqrt{x^2 - 4x + 5}} \cdot (2x - 4) = 0 $$
$$ 2x - 4 = 0 $$
$$ x = 2 $$

3. Check for points where the derivative is undefined:
The derivative is undefined when the denominator of $ f'(x) $ equals zero:
$$ x^2 - 4x + 5 = 0 $$
$$ (x - 2)^2 + 1 = 0 $$

The above equation has no real solutions.

Thus, the only critical point of $ f(x) $ is $ x = 2 $.