VCE Stuff > VCE Mathematical Methods CAS

Question..

(1/2) > >>

hifer:
Given f(x) = (x-1)^2 and g(x) = sqrt(x) + 1, find g(f(x))

Somehow the soln says that u can't simplify sqrt((x-1)^2) to just x-1 ? Why is that?

reg:
Because sqrt is really +-. The full graph creates an absolute value graph, because you can't have a negative when you square something.

f(x) = sqrt((x - 1) ^ 2)
g(x) = (x - 1)

f(-1) = sqrt((-2) ^2) = sqrt(4) = 2
g(-1) = -2

Fitness:

--- Quote from: "reg" ---Because sqrt is really +-. The full graph creates an absolute value graph, because you can't have a negative when you square something.
--- End quote ---

Wrong. You can't have sqrt(-ve number). You can have -sqrt(number). (-ve)^2 does exist as well. So I don't really see your point there ...
And it doesn't say +-, It defines it as only +...

It may say the answer is sqrt((x-1)^2) but that could be because they don't expect you to simplify? Are there any comments with that answer?

reg:
Er, excuse my poor expression perhaps, but any (-ve)^2 will be positive, which is what I meant.

It's sort of like when you take an inverse of a parabola:
y = x^2
y = sqrt(x)

You are only getting the +ve because you cannot get a negative value unless  you explicitly +- the function.

With the OP's function, you have to square and sqrt separately because the ^2 operation means only positive values will exist.

hifer:

--- Quote from: "reg" ---Because sqrt is really +-. The full graph creates an absolute value graph, because you can't have a negative when you square something.

f(x) = sqrt((x - 1) ^ 2)
g(x) = (x - 1)

f(-1) = sqrt((-2) ^2) = sqrt(4) = 2
g(-1) = -2
--- End quote ---


I've plot the graph and see ur point that it's a absolute function. the answers says u either leave it in sqrt((x-1)^2) form or |x-1| form.. How do u identify that it's an absolute function?

I played around on my calc and see that whenever there's a sqrt(x^2) eqn, it's an absolute function, is that how u identify it?

Navigation

[0] Message Index

[#] Next page