Can a Square Root Function Have Two Slopes and Exist in the Fourth Quadrant?

[Rodrae]

y = [itex]\sqrt{x^2- 2x + 1}[/itex]
The y' is always the slope right?
Then if we simplify the equation
y =[itex]\sqrt{(x-1)^2 }[/itex]----or---y = [itex]\sqrt{(1-x)^2}[/itex]
y = x - 1 y = 1-x
Checking:---------------Checking:
[itex]\sqrt{(x-1)^2}[/itex]-------------[itex]\sqrt{(1-x)^2}[/itex]
[itex]\sqrt{x^2-2x+1}[/itex]-----------[itex]\sqrt{1-2x + x^2}[/itex]

y'= -----------------------------------y' = -1

Using the law of derivatives then
[itex]\frac{2x-2}{2\sqrt{x^2-2x+1}}[/itex]
and simplifying this will also gave 2 answers...


So the problem is:
Is there a posibility that there will be 2 y' ?
And also the graph of y=[itex]\sqrt{x}[/itex]
Is always in the first quadrant but don't you think it could be in the 4th quadrant because if x = 1 then y = 1 and y=-1
(-1)(-1) = 1 and (1)(1) = 1

 

[Hootenanny]

Perhaps taking a look at a plot of the function [itex]f(x) = \sqrt{x^2-2x+1}[/itex] will help you:http://www.wolframalpha.com/input/?i=sqrt{x^2-2x%2B1}

When we write [itex]\sqrt{\cdot}[/itex], we mean what it called the principle square root, which is defined such that it is always positive. This helps avoid confusing scenarios such as this. In other words, we define [itex]f(x)[/itex] in a piecewise fashion:

[tex]f(x) = \sqrt{x^2-2x+1} :=\left\{\begin{array} x x-1 & x > 1 \\ 1-x & x < 1\end{array}\right.[/tex]

See the wikipedia page and in particular the definition of the square root of [itex]x^2[/itex]: http://en.wikipedia.org/wiki/Square_root#Properties

 

[alexfloo]

The only spot you went wrong is when you assumed that

[itex]\sqrt{x^2}=x[/itex].

In fact,

[itex]\sqrt{x^2}=|x|[/itex].

Since the absolute value can be considered as piecewise-defined, this is what leads to the piecewise derivative that Hootenany gave.

 

[jackmell]

We're limiting our interpretation of the root function by restricting it to it's principal value. This guy, Rodrae, is really anticipating the behavior of the complex counterpart of the root function (when the arguments are complex variables): it does have two values, and it's derivative has two values as well. y' is not always the slope, rather, it's the derivative of y and if y is multivalued such as the square root, then the derivative is multivalued as well. If you drew plots of the complex functions, you'd see why it and it's derivative has two values. The confussion in my opinion comes from our restrictions we place on these multifunctions until the student is in college studying complex variables and is shocked by the concept of "multifunction" and heaven forbid, that of a "branch-cut". Then they are further confussed by the current inability of most classes and textbook, in my opinion, to present this subject well.

So if I may be allowed to disagree with others in this post and answer the thread author's questions, yes it does have two values and so does it's derivative.

 

[simplicity123]

y = [itex]\sqrt{x^2- 2x + 1}[/itex]
The y' is always the slope right?
Then if we simplify the equation
y =[itex]\sqrt{(x-1)^2 }[/itex]----or---y = [itex]\sqrt{(1-x)^2}[/itex]
y = x - 1 y = 1-x
Checking:---------------Checking:
[itex]\sqrt{(x-1)^2}[/itex]-------------[itex]\sqrt{(1-x)^2}[/itex]
[itex]\sqrt{x^2-2x+1}[/itex]-----------[itex]\sqrt{1-2x + x^2}[/itex]

y'= -----------------------------------y' = -1

Using the law of derivatives then
[itex]\frac{2x-2}{2\sqrt{x^2-2x+1}}[/itex]
and simplifying this will also gave 2 answers...


So the problem is:
Is there a posibility that there will be 2 y' ?
And also the graph of y=[itex]\sqrt{x}[/itex]
Is always in the first quadrant but don't you think it could be in the 4th quadrant because if x = 1 then y = 1 and y=-1
(-1)(-1) = 1 and (1)(1) = 1

y=|x-1|

The rest is just nonsense. Also, you can't differentiate that function. y=|x-1| isn't differentiable as it's not continuous at x=1.

 

[Dr. Seafood]

Also, you can't differentiate that function. y=|x-1| isn't differentiable as it's not continuous at x=1.

Actually, you can differentiate it everywhere except at x = 1. The function is continuous everywhere, just not differentiable at that particular number.

 

[alexfloo]

You'll notice that when you simplify the one you got by chain rule, you'd find [itex]\pm 1[/itex] are the answers. The original function you had (technically, using the interpretation of the square root as the principal root) is the function y=x. If we plot *both* roots, however, you have the function y=x and y=-x, which is why you get two slopes, 1 and -1. They correspond to the two "branches" of the function.

 

[pwsnafu]

This guy, Rodrae, is really anticipating the behavior of the complex counterpart of the root function (when the arguments are complex variables):

Disagree on this. The OP contains

The y' is always the slope right?

<snip>

And also the graph of y=[itex]\sqrt{x}[/itex]
Is always in the first quadrant but don't you think it could be in the 4th quadrant

These are concepts valid in real analysis, not complex analysis.

 

[Hootenanny]

You'll notice that when you simplify the one you got by chain rule, you'd find [itex]\pm 1[/itex] are the answers. The original function you had (technically, using the interpretation of the square root as the principal root) is the function y=x. If we plot *both* roots, however, you have the function y=x and y=-x, which is why you get two slopes, 1 and -1. They correspond to the two "branches" of the function.

Just to clarify for the OP, the principle square root of [itex]x^2[/itex] is not [itex]x[/itex], as suggested here. It is the piece-wise function [itex]|x|[/itex]. Notice that although the codomain of the square root is [itex]\mathbb{R}^+[/itex], its domain is [itex]\mathbb{R}[/itex].

Just to reiterate, unless [itex]f(x)[/itex] is non-negative, it is incorrect in general to say that [itex]\sqrt{f^2(x)} = f(x)[/itex], rather [itex]\sqrt{f^2(x)} = |f(x)|[/itex] or

[tex]\sqrt{f^2(x)}=\left\{\begin{aligned} f(x) & \text{for} & f(x)\geq 0 \\ -f(x) & \text{for} & f(x)<0\end{aligned}\right..[/tex]

 

[HallsofIvy]

y=|x-1|

The rest is just nonsense. Also, you can't differentiate that function. y=|x-1| isn't differentiable as it's not continuous at x=1.

y= |x- 1| is continuous at x= 1 and is differentiable for all x except 1.

 

[Rodrae]

So therefore you mean that ...
for example y= [itex]\sqrt{(cos180)(sin270)}[/itex] = [itex]\sqrt{(-1)(-1)}[/itex] = [itex]\sqrt{(-1)^2}[/itex]
and we can't say (-1)^2(1/2)
= (-1)^2/2
= (-1)¹
= -1


and you say that it is absolute value so it will =1 and will never be equal to -1?

and[itex]\sqrt{x^2}[/itex] is just like |x|
so therefore[itex]\sqrt{}[/itex] nos will never be negative?

 

[Hootenanny]

So therefore you mean that ...
for example y= [itex]\sqrt{(cos180)(sin270)}[/itex] = [itex]\sqrt{(-1)(-1)}[/itex] = [itex]\sqrt{(-1)^2}[/itex]
and we can't say (-1)^2(1/2)
= (-1)^2/2
= (-1)¹
= -1

No, you can't say that. You switched branches midway through your computation.

and[itex]\sqrt{x^2}[/itex] is just like |x|
so therefore[itex]\sqrt{}[/itex] nos will never be negative?

Without qualification, no. If you want to denote the negative root of something then you must put [itex]-\sqrt{\cdot}[/itex]. On its own, [itex]\sqrt{\cdot} \geq 0[/itex].

 

[Rodrae]

ok. Thanks a lot