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Indicate a method to find the inverse of the function f(x)=square root(x-2)/5?
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Let f(x) = y = [sqrt (x - 2)]/5
express x in terms of y
y = [sqrt (x - 2)]/5
=> 5y = sqrt (x - 2)
=> 25y^2 = (x - 2)
=> x = 25*y^2 + 2
interchange x and y
=> y = 25*x^2 + 2
The inverse of f(x) , f^-1(x) = 25x^2 + 2
Posted by justaguide on May 14, 2011 at 4:10 PM (Answer #1)
Whenever we want to determine the inverse function, we'll have to write x with respect to y.
Let f(x) = y and we'll re-write:
y = sqrt(x-2)/5
We'll multiply both sides by 5:
5y = sqrt(x-2)
We'll raise to square both sides, to eliminate the square root:
(5y)^2 = x-2
We'll use the symmetric property:
x-2 = 25y^2
We'll add 2 both sides:
x = 25y^2 + 2
Therefore, the required inverse function is: f^-1(x) = 25x^2 + 2.
Posted by giorgiana1976 on May 14, 2011 at 4:17 PM (Answer #2)
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