Calculus I |
Quiz # 7
|
November 10, 2003
|
Name____________________ |
Score ______
|
R. Hammack
|
(1) Find the inverse of the function
.
![y = 3x^3 - 4](HTMLFiles/quiz7sol_2.gif)
![y + 4 = 3x^3](HTMLFiles/quiz7sol_3.gif)
![(y + 4)/3 = x^3](HTMLFiles/quiz7sol_4.gif)
![(y + 4)/3^(1/3) = x](HTMLFiles/quiz7sol_5.gif)
![y = (x + 4)/3^(1/3)](HTMLFiles/quiz7sol_6.gif)
![f^(-1)(x) = (x + 4)/3^(1/3)](HTMLFiles/quiz7sol_7.gif)
(2) Consider a one-to-one differentiable
function
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](HTMLFiles/quiz7sol_9.gif)
(a) ![f^(-1)(3) =](HTMLFiles/quiz7sol_10.gif)
From the table, you see that
.
Thus, you get
,
and this is
,
Therefore ![f^(-1)(3) = 2](HTMLFiles/quiz7sol_14.gif)
(b) If
,
find
.
![g ' (x) = d/dx[f^(-1)(x) + 8] = d/dx[f^(-1)(x)] = 1/(f ' (f^(-1)(x)))](HTMLFiles/quiz7sol_17.gif)
Thus ![g ' (3) = 1/(f ' (f^(-1)(3))) = 1/(f ' (2)) = 1/5](HTMLFiles/quiz7sol_18.gif)
(3) Simplify the following
expressions as much as possible.
(a) ![log(100^(1/3)) =](HTMLFiles/quiz7sol_19.gif)
![log_10(10^(2/3)) = 2/3](HTMLFiles/quiz7sol_20.gif)
(b) ![FormBox[RowBox[{RowBox[{(-32), ^, RowBox[{(, RowBox[{-, 0.6}], )}]}], =}], TraditionalForm]](HTMLFiles/quiz7sol_21.gif)
![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]](HTMLFiles/quiz7sol_22.gif)
(c) ![log(25) + 2 log(2) =](HTMLFiles/quiz7sol_23.gif)
![log(25) + log(2^2) = log(25) + log(4) = log(100) = 2](HTMLFiles/quiz7sol_24.gif)
(d)
0
(e)
4