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Determine the inverse g(x) of the function f(x) = (1+(4/x)), stating its domain and range. Verify that f(g(x)) = g(f(x))=x and that g′(f(x)) = (1/(f′(x))?
Chin Chin
Let f(x) = y
y = 1 + (4/x)
Now replace y with x and x with y and find equation for y
x = 1 + (4/y)
(x-1) = (4/y)
y = 4/(x-1)
This g(x), the inverse of f(x)
g(x)= 4/(x-1)
The domain will be all real numbers except when (x-1)=0 or x=1
So Domain = (-∞,1),(1,+∞)
And Range = (-∞,0),(0,+∞)
f(g(x)) = f(4/(x-1)) = 1 + 4/(4/(x-1)) = 1+(x-1) = x
g(f(x)) = g(1+(4/x)) = 4/((1+(4/x))-1) = 4/(4/x) = x
So we get f(g(x)) = g(f(x))
Notice the error in copying the next part of your question
It should be g'(f(x)) = 1/(f'(g(x)))
g'(f(x)) = d/dx (g(f(x))) = d/dx (x) = 1
f'(g(x)) = d/dx (f(g(x))) = d/dx (x) = 1
1/[f'(g(x))] = 1/1 = 1
g'(f(x)) = 1/f'(g(x)) ( Notice the error in copying your question)
Ash
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