ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: jag on February 10, 2009, 07:02:12 pm
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The functions f and g are defined by f: R -> R , f(x) = x2-x and g:{x:x greater than or equal to 0} --->R, where g(x) = root of x.
(a) Explain why g o f does not exist
(b) find f o g
note: how do i do square root or greater than or equal to sign on this forum. ?
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a) g o f does not exist because the range of f is not a subset of the domain of g.
If you think about it, the range of x^2 - x is [-1/4, infinity).
You're then subbing f into g, so instead of sqrt(x) it's now sqrt(g) and if g has negative values, then it doesn't exist, because you cannot square root negative numbers in the real domain.
b) f o g = f(g(x)) = (sqrt(x))^2 - sqrt(x) = x - sqrt(x)
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the answer to (a) in the book says:
f of g is not defined since ran g = [-1,infinty) is not a subset of dom f = (-infinity,3]
could som1 please explain what this all is as to why the range of g is what it is and how the dom of f is what it is?
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the book is wrong... or you're reading from the wrong answer.
if you think about it,
it's essentially saying that there's a value for which x^2 - x = -1
so x^2 - x + 1 = 0
using quadratic formula, x = (-(-1) +/- sqrt((-1)^2 - 4(1)(1))/2(1)) = (1 +/- sqrt(-3))/2
you can't sqrt a negative number, so, the range from the book is wayyyyy wrong.
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oh sorrrry the f function is actually f(x) = x^2 - 2 in the question.
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note: how do i do square root or greater than or equal to sign on this forum. ?
http://vcenotes.com/forum/index.php/topic,3137.0.html