Calculus

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)

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  1. 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)

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  2. Thank you so much!

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