If f(x)=4x-(1/x) when x>0 and g is the inverse of f, then what is the value of g'(0)?

Answer 1

f(x)=4x-(1/x)
or
y = 4x - 1/x

step 1 of inverse: interchange the x and y's
x = 4y - 1/y
multiply by y
xy = 4y^2 - 1 <----- the g relation
when x = 0, 0 = 4y^2 - 1 ---> y = 1/2
xy' + y = 8y y'
xy' - 8y y' = -y
y'(x-8y) = -y
y' = -y/(x-8y) = y/(8y - x) <----- g'
plug in x=0, y=1/2
g'(0) = (1/2) / (8(1/2))
= 1/8