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Calculus I                                                          Quiz #8                            November 9, 2004

Name____________________                   R.  Hammack                              Score ______
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(1)  Suppose an invertible function f and its derivative obey the following table. x f(x) f ' (x) 0 4 -2 1 3 0 2 1 -1 3 0 -2
(a)  f^(-1)(0) =3    (because f(3) = 0)

(b)    Is f ' (x)invertible?  Explain.    NO. The chart says f '(0) = f '(3),  so f '  is not one-to-one, so it's not invertible.


(c) If g(x) = f^(-1)(x),  find g ' (1).

Using the formula for the derivative of an inverse, we get:
g ' (x) = d/dx[f^(-1)(x)] = 1/(f ' (f^(-1)(x)))
Thus:
g ' (1) = 1/(f ' (f^(-1)(1))) = 1/(f ' (2)) = 1/-1 = -1



(2)   Find the inverse of the function  f(x) = 3/(1 + x) + 5

y = 3/(1 + x) + 5  y - 5 = 3/(1 + x)  (y - 5) (1 + x) = 3  y + y x - 5 ... y  x = (8 - y)/(y - 5)  y = (8 - x)/(x - 5)  f^(-1)(x) = (8 - x)/(x - 5)


(3)   The graph of a function  f  is drawn. Sketch the graph of its inverse.

The graph is obtained by reflecting the given graph across the line y = x.

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