don't understand this math function: f: xa *square root of (3-2x)
I can only recognize functions like, eg. f: x = x²
what is the 'a' doing there?
I have to evaluate f^-1(5)
f(x)=a√(3-2x)...x≦3/2
y=a√(3-2x)
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ã ã if a＞0 then y≧0
ã ã if a＜0 then y≦0
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x=a*√(3-2y)...y≦3/2
ã ã if a＞0 then x≧0
ã ã if a＜0 then x≦0
x^2=(a^2)(3-2y)
[(x^2)/(a^2)]=3-2y
2y=3-[(x^2)/(a^2)]
y=(3/2)-[(x^2)/2(a^2)]
ã f^-1_(x)=(3/2)-[(x^2)/2(a^2)]
ã f^-1_(5)=(3/2)-[25/2(a^2)]ã .
f:xa = √(3-2x)
You can convert this into the form that you recognize with the introduction of a new variable.
Let X = xa
Then, x = X/a
So,
f:X = √(3-2(X/a))
Maths prob i dont understand algebra can sum 1 explane it 2 me::
8 posts - Last post: Feb 22, 2008Maths prob i dont understand algebra can sum 1 explane it 2 me You asked about a (superscript 2) + c (superscript 2) Which in text is writen
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f:X = √(3 - 2X/a)
Now you can find your inverse function.
X = √(3 - 2f^-1(X)/a)
X^2 = 3 - 2f^-1(X)/a
2f^-1(X)/a = 3 - X^2
f^-1(X) = (a/2)(3 - X^2)
f^-1(5) = (a/2)(3 - 25) = (a/2)(-22) = -11a
Or, you could find the answer directly without actually finding the inverse function this way:
5 = √(3 - 2x)
25 = 3 - 2x
2x = -22
x = -11
So, f(-11a) = 5
Therefore, f^-1(5) = -11a
EDIT..........post Additional Details
Perhaps there was a misprint. Usually, (and it appears as if you are British, so I hope this is not misleading, as I did not study there, but mathematics should be consistant worldwide, and a function definition can be written as f:x ) , a function is written as:
f:x → (some function of x)
or
f(x) = (some function of x)
Was there really a " * " involved?
Maybe this should have been an equal sign, or an arrow which was not displayed properly by computer, or some other format of function definition given in your course of study.
I had originally assumed this might have been a typo.
a is a constant of indeterminate value. a will appear in your answer.
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