Inverse Functions

I.  Quiz

II.  Homework

III.  Inverse Functions (Definition)

    Let f(x) be a 1-1 function then g(x) is an inverse function of f(x) if

        f(g(x)) = g(f(x)) = x

Example:

    For f(x) = 2x - 1, f^-1(x) = 1/2 x +1/2

    Since f(f^-1(x) ) = 2[1/2 x +1/2] - 1 = x

    and f^-1(f(x)) = 1/2 [2x - 1] + 1/2 = x

IV.  The Horizontal Line Test and Roll's Theorem

    Note that if f(x) is differentiable and the horizontal line test fails then f(a) = f(b) and Rolls theorem implies that there is a c such that f'(c) = 0.  A partial converse is also true:  

    If f is differentiable and f'(x) is always non negative (or  always non positive) then f(x) has an inverse.

Example:

    f (x) = x³ + x - 4 has an inverse since

    f'(x) = 3x² + 1 which is always positive.

V.  Continuity and Differentiability of the Inverse Function 

Theorem:

1. f continuous implies that f^-1 is continuous.

2. f increasing implies that  f^-1 is increasing.

3. f decreasing implies that f^-1 is decreasing

4. f differentiable at c and f'(c)  is not 0 implies that f^-1 is differentiable at f(c).

5. If g(x) is the inverse of  the differentiable f(x) then 

   g'(x) = 1/[f'(g(x))]

    if f'(g(x)) not 0.

    

Proof  Since f(g(x)) = x we differentiate implicitly:

    d/dx[f(g(x)) ] = d/dx [x]

    Using the chain rule y = f(u),  u = g(x)

    dy/dx = (dy/du)(du/dx) = (f'(u))(g'(x)) = (f'(g(x)))(g'(x))

    So that

    (f'(g(x)))(g'(x)) = 1  Dividing, we get:

    g'(x) = 1/(f'(g(x)))

Example:  

    For x > 0 Let f(x) = x² and g(x) = sqrt(x) be its inverse, then

    g'(x) = 1/[2(sqrt(x)]  

    Note that d/dx [sqrt(x)] = d/dx[x^1/2] = 1/2 x^-1/2  = 1/[2sqrt(x)]

Exercise:  

    A)   Let f(x) = x³ + x - 4

            Find d/dx[f^-1(-4)]

    B)  Let f(x) = int from 2 to x of 1/(1 + x³)dx

            Find  d/dx[f^-1(0)]