ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: hifer on November 08, 2007, 02:31:45 pm
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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?
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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
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Because sqrt is really +-. The full graph creates an absolute value graph, because you can't have a negative when you square something.
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?
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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.
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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
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?
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Er, excuse my poor expression perhaps, but any (-ve)^2 will be positive, which is what I meant.
Sorry, misread.
This is a clever question.
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Absolute value function can be identified when you have a sqrt (something) squared.
E.g sqrt(9x+1)squared is equal to abs(9x+1)
This is because a square root can only produce positive real values, just like an abs function.
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I dno about specific technique, but if a function exists x < 0, and x is being squared, you are losing its sign, it will be positive regardless.
With your function you can see how sqrt(x ^ 2) can only ever be positive, it's , as you say, equal to |x|.
x on its own is equal to both the -ve and +ve values.
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yup, got that now thx guys
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I dno about specific technique, but if a function exists x < 0, and x is being squared, you are losing its sign, it will be positive regardless.
With your function you can see how sqrt(x ^ 2) can only ever be positive, it's , as you say, equal to |x|.
x on its own is equal to both the -ve and +ve values.
When any real number is squared it becomes positive. However, the square root of any positive number could be positive or negative. Going in the reverse again, "all real numbers" includes both positive and negative values. So (-3)^2 = (3)^2 = 9
And root(9) = +/-3
Hope that helps