f(x) = √(2x)
g(x) = f^-1(x) = x^2/2
f'(x) = 1/√(2x)
g'(x) = x
So, we want
1/√(2c) = c
1/2c = c^2
1/2 = c^3
c = (1/2)^(1/3)
Suppose that y=f(x) = sqrt(2x), x>=0
Find a c > 0 such that the tangent line to the curve y = f(x) at x = c has the same slope as the tangent line to the curve y = f^–1(x) at x = c.
You get:
c = 1/8
c = 1/2
c = (1/8)^(1/3)
c = (1/3)^(2/3)
c = (1/2)^(1/3)
2 answers
Thank you so much!