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Calculus I                                            Test #3                             May 4, 2004

Name__________________           R.  Hammack                   Score ______
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(1)  Sketch the graph of the function f(x) = e^(x - 1) - 2.
The function y = e^xis drawn dashed.
The function y = e^(x - 1)is the previous graph shifted one unit right. It's drawn dotted.
The function y = e^(x - 1) - 2is the previous graph shifted 2 unit down. It's drawn solid.

[Graphics:HTMLFiles/T3S04C_6.gif]

(2) Find the inverse of the function f(x) = (x + 1)/(x - 1).

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

(3) Does the function  g(x) = 2cos(x) + x  have an inverse?  Explain.

Notice that  g ' (x) = -2sin(x) + 1 can have either positive or  negative values, depending on x.
Examples:
g ' (0) = -2sin(0) + 1 = 1>0
g ' (-π/2) = -2sin(π/2) + 1 = -1<0

Thus the function g increases and decreases. It fails the Horizontal Line Test.
There is no inverse.


(4) Solve the equation   4 · 5^(x + 1) = 3.  (It's OK to have logarithms  in your final answer.)

5^(x + 1) = 3/4  ln(5^(x + 1)) = ln(3/4)  (x + 1) ln(5) = ln(3/4)  x l ... 62371; x ln(5) = ln(3/4) - ln(5)  x = (ln(3/4) - ln(5))/ln(5)  x = ln(3/20)/ln(5)

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

(a)  FormBox[RowBox[{RowBox[{(-32), ^, 0.2}], =}], TraditionalForm](-32)^(1/5) = (-32)^(1/5) = -2

(b) e^(2ln(3) + ln(5)) = e^(ln(3^2) + ln(5)) = e^(ln(9) + ln(5)) = e^ln(45) = 45

(c)  tan^(-1)(-3^(1/2)) =-π/3

(d)  cos^(-1)(cos((5π)/3)) =cos^(-1)(1/2) = π/3



(6)  Find the following derivatives.

(a)   d/dx[ e^(x ln(x) )] =e^(x ln(x) )((1) ln(x) + x1/x) = e^(x ln(x) )(ln(x) + 1)

(b)   d/dx[   (  ln ( tan^(-1)(x) )   )^3 ] =3 (ln ( tan^(-1)(x) ))^2d/dx[  ln ( tan^(-1)(x) )   ] = 3 (ln ( tan^(-1)(x) ))^21/(1 + x^2)/tan^(-1)(x) = (3 (ln ( tan^(-1)(x) ))^2)/(tan^(-1)(x) (1 + x^2))


(c)   d/dx[  log_10(x) ] =1/(ln(10) x)


(d)    d/dx[sin^(-1)(x^(1/2))] =1/(1 - x^(1/2)^2)^(1/2) 1/2x^(-1/2) = 1/(2x^(1/2) (1 - x)^(1/2))


(7) The graph of the derivative of a function f(x) is drawn. Answer the following questions about the function f(x).

[Graphics:HTMLFiles/T3S04C_31.gif]
(a) Find the interval(s) on which f(x) increases.
f(x)  INCREASES on (-∞, 0] and [2,3], because that's where its derivative is positive.

(b)  Find the critical points of f(x).
-3, 0, 2, 3

(c)  Find the locations of the relative extrema of f(x), and identify them as relative maxima or minima.
Relative Max at 0 and 3 (Derivative switches from + to -)
Relative Min at 2 (Derivative switches from - to +)

(d)  Find the intervals on which the graph of f(x) is concave up/down.
CONCAVE UP on [-3 ,-1] & [1, 2.5] (That's where f  ' increases, so f " is positive.)
CONCAVE DOWN on (-∞ , -3] & [-1, 1] & FormBox[RowBox[{[, RowBox[{2.5, ,, ∞}], )}], TraditionalForm] (That's where f  ' decreases, so f " is negative.)


(8) These problems  are about the function f(x) = 3x^4 - 4x^3. Answer the following questions about  f(x).

(a) Find the critical points of f(x).
f ' (x) = 12x^3 - 12x^2 = 12x^2(x - 1) = 0
The critical points (which make the derivative 0) are 0 and 1

(b)  Find the intervals of increase/decrease of f(x).
     0     1      
----+---+-----
- - - - -  | + + f '(x)

f increases on [1, ∞)
f decreases on (-∞, 1]


(c)  Find the locations of the relative extrema of f(x), and identify them as relative maxima or minima.
Relative minimum at x = 1,
No relative maximum.


(d)  Find the intervals on which the graph is concave up/down.

f ' (x) = 12x^3 - 12x^2 f '' (x) = 36x^2 - 24x = 12x(3x - 2)

     0     2/3      
----+---+-----
+ +| - - | + + f ''(x)

f concave up on (-∞, 0],[2/3, ∞)
f concave down on [0, 2/3]