- #1
Rodrae
- 13
- 0
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
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