Quick simple question - What is the difference between to graphs?? (1 Viewer)

[blackops23]

Ok, f(x)=[(x-1)^2]/[x+1]

g(x) = [x-1]/[(x+1)^0.5]

h(x) = +sqrt(f(x))



What is the difference between g(x) and h(x)? At first I thought g(x) = h(x), but after sketching them on a graphing software, they weren't the same, infact it looked more like h(x) = |g(x)| i.e. - (h(x) is the absolute value of h(g)...)

I was confused because I had to sketch g(x) from f(x), and I thought all you had to do was (sqrt) the y-coordinates of f(x), (therefore, g(x) will be positive for all x) but then the y-intercept of g(x) was (0,-1) -- A NEGATIVE!!

I then "graphed" h(x) on Graphmatica, and indeed it held true to it being POSITIVE for all x, so my question:

Why isn't g(x) = +sqrt(f(x)), why are there negatives on the graph of g(x)??

Thanks guys, sorry for rambling...

 

[blackops23]

I spelt "two" wrong in the thread title.

FML.

 

[deterministic]

+sqrt(f(x)) must be greater than or equal to 0 for ALL x.

So:
+sqrt(f(x))= +sqrt((x-1)^2))/(x+1)^0.5
(x+1)^0.5 is always positive by definition

Consider +sqrt((x-1)^2) <-- this must be greater than or equal to 0 for ALL x.
So if x-1>=0, then +sqrt((x-1)^2)=x-1
If x-1=<0, then 1-x>=0 and hence +sqrt((x-1)^2)=1-x

FOR x-1>=0 (ie. x>=1) => +sqrt(f(x)) = g(x)
FOR x-1<=0 (ie. x<=1) => +sqrt(f(x))= (1-x)/(x+1)^0.5 = -g(x)

So indeed sqrt(f(x))=|g(x)|

 

[blackops23]

+sqrt(f(x)) must be greater than or equal to 0 for ALL x.

So:
+sqrt(f(x))= +sqrt((x-1)^2))/(x+1)^0.5
(x+1)^0.5 is always positive by definition

Consider +sqrt((x-1)^2) <-- this must be greater than or equal to 0 for ALL x.
So if x-1>=0, then +sqrt((x-1)^2)=x-1
If x-1=<0, then 1-x>=0 and hence +sqrt((x-1)^2)=1-x

FOR x-1>=0 (ie. x>=1) => +sqrt(f(x)) = g(x)
FOR x-1<=0 (ie. x<=1) => +sqrt(f(x))= (1-x)/(x+1)^0.5 = -g(x)

So indeed sqrt(f(x))=|g(x)|

So... is it safe to say that sqrt(f(x)) = |g(x)| wher g(x) is *something* of/to f(x)... I got no idea. could someone please give me a link or a rule that connects the sqrtf(x) to the absolute value of the sqrt... I think I'm getting a bit confused...

 

[deterministic]

sqrt(f(x)) by definition is always positive, so if g(x)^2=f(x), then sqrt(f(x))=|g(x)|

Consider a numerical example:
2^2=4 also (-2)^2=4
but sqrt(4)=2 = |-2| =|2|

rather g(x)=+sqrt(f(x)) for some values of x (in this case x>=1), -sqrt(f(x)) for other values of x (in this case x<=1)