Calculus I
Quiz # 7
November 10, 2003
Name____________________
Score ______
R.  Hammack 


(1) Find the inverse of the function f(x) = 3x^3 - 4.

y = 3x^3 - 4

y + 4 = 3x^3

(y + 4)/3 = x^3

(y + 4)/3^(1/3) = x

y = (x + 4)/3^(1/3)

  f^(-1)(x) = (x + 4)/3^(1/3)


(2) Consider a one-to-one differentiable function f(x) which satisfies the following table.    x 1 2 3 4 5 f(x) 0 3 6 8 9 f ' (x) 1 5 3 3 2
(a) f^(-1)(3) =

From the table, you see that f(2) = 3.
Thus, you get f^(-1)(f(2)) = f^(-1)(3),
and this is  2 = f^(-1)(3),
Therefore  f^(-1)(3) = 2


(b) If g(x) = f^(-1)(x) + 8, find g ' (3).

g ' (x) = d/dx[f^(-1)(x) + 8] = d/dx[f^(-1)(x)] = 1/(f ' (f^(-1)(x)))
Thus g ' (3) = 1/(f ' (f^(-1)(3))) = 1/(f ' (2)) = 1/5



(3)  Simplify the following expressions as much as possible.

(a)   log(100^(1/3)) =log_10(10^(2/3)) = 2/3

(b)   FormBox[RowBox[{RowBox[{(-32), ^, RowBox[{(, RowBox[{-, 0.6}], )}]}], =}], TraditionalForm]FormBox[RowBox[{StyleBox[RowBox[{1, /, RowBox[{(-32), ^, 0.6}]}], FontSize -> 16], =, 1/(-32)^(3/5) = 1/((-32)^(1/5))^3 = 1/(-2)^3 = -1/8}], TraditionalForm]

(c)    log(25) + 2 log(2) =log(25) + log(2^2) = log(25) + log(4) = log(100) = 2

(d)   log_3(1) =0

(e) 3^log_3(4) =4