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Calculus I                                                       Test #3                                       November 25, 2002

Name____________________                R.  Hammack                                         Score ______
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(1) The following function is one-to-one. Find its inverse.  f(x) = e^( x^3 + 1)
y = e^( x^3 + 1)
x = e^( y^3 + 1)
ln(x) = ln (e^( y^3 + 1) )
ln(x) = y^3 + 1
 y^3 = ln(x) - 1  
 y = (ln(x) - 1  )^(1/3)
FormBox[RowBox[{ , RowBox[{RowBox[{f^( -1)(x), Cell[]}], =, (ln(x) - 1  )^(1/3)}]}], TraditionalForm]

(2) Decide if the function  f(x) = e^x + 50x is invertible.  Explain your reasoning.
Notice that   f ' (x) = e^x + 50 is positive no matter what value x has. This means the graph of f  increases, and never decreases. Thefore f passes the horizontal line test and consequently is invertible.


(3)   Solve the equation   7 = 5e^(x + 2)   for x.
ln(7) = ln(5e^(x + 2))  
ln(7) = ln(5) + ln ( e^(x + 2))  
ln(7) - ln(5) = x + 2    
x = ln(7) - ln(5) - 2    
x = ln(7/5) - 2
(4) Simplify the following expressions as much as possible.

(a)    8^(-4/3) =(8^(1/3))^(-4) = 2^(-4) = 1/2^4 = 1/16

(b)     e^(ln(3) + ln(2)) =   e^ln(3  2) = e^ln(6) =6

(c)     log_2(4^(-3/2)) = -3/2log_2(4) = -3/2(2) = -3

(d)     ln(1/e^9) = -9

(e)     ln(25) + 2 ln(e/5) = 2ln(5) + 2 ln(e/5) =2 (ln(5) + ln(e/5)) =2 ln((5e)/5) =2 ln(e) =2

(f)   sin^(-1)(-1/2) =-π/6

(g)    sin ( cos^(-1)(x)) =(1 - x^2)^(1/2)

  
  
(5) The graph of the derivative f ' (x) of a function f(x) is sketched.

[Graphics:HTMLFiles/3F02_40.gif]

Answer the following questions about f(x).

(a)  Find the interval(s) on which  f  is increasing.
This happens on the intervals on which f '(x) is positive, namely (-1, 3) and (3, ∞).

(b)  Find the interval(s) on which  f  is concave down.
This happens on the intervals on which f ''(x) is negative.
That happens on the intervals on which f '(x) is decreasing, namely (-5, -3) and (1, 3).

(c)  List the x-coordinates of all inflection points of  f.
These will be where f ''(x) changes sign, or, in terms of f '(x), where f '(x) changes from increasing to decreasing or decreasing to increasing. Namely: -5, -3, 1 and 3

(d)  List all the critical numbers of f.
These are where  f '(x) = 0, and from the graph they are -5, -1 and 3

(e)  List the x-coordinates of the relative minima of  f  (if any).
Only at x = -1, because that's where f '(x) switches from negative to positive.


(6) Find the following derivatives.
(a)   d/dx[  e^(1 - x e^(-x)) ] =   e^(1 - x e^(-x)) (0 - e^(-x) - x e^(-x)(-1)) =  e^(1 - x e^(-x)) (x e^(-x) - e^(-x))

(b)   d/dx[  tan^(-1)(x^3)   ] =1/(1 + (x^3)^2) 3x^2 =(3x^2)/(1 + x^6)

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



(d)  d/dx[  x^( cos(x)) ]= x^( cos(x))(cos(x)/x - sin(x) ln (x ))   (as computed below)

We use logarithmic differentiation.
y =   x^( cos(x))

ln(y) =   ln (x^( cos(x)) )

ln(y) = cos(x) ln (x )

d/dx[ln(y)] = d/dx[ cos(x) ln (x )]

1/ydy/dx = -sin(x) ln (x ) + cos(x) 1/x

dy/dx = y(-sin(x) ln (x ) + cos(x) 1/x)

dy/dx = x^( cos(x))(cos(x)/x - sin(x) ln (x ))


(7) Consider the function f(x) = ln(x^4 - 2x^2 + 20)

(a) Find all the critical numbers of f.

f ' (x) = (4x^3 - 4x)/(x^4 - 2x^2 + 20)= (4x(x^2 - 1))/(x^4 - 2x^2 + 20)= (4x(x - 1) (x + 1))/(x^4 - 2x^2 + 20)

Thus the ciritical numbers of f are -1, 0 and 1 because they make the derivative zero.


(b) Find the locations of all the relative extrema of f, and classify them as relative maximums or minimums.
         -1     0       1
------+-----+-----+-------
- - - -| + + | - - - | + + + f '(x)

By the first derivative test:
f has a local maximum at x = 0.
f has local minima at x = -1 and x = 1

(8) A formula from physics states that an object which is propelled up or down with an initial velocity of v_0feet per second from a height of s_0feet has a height of  s(t) = s_0 + v_0t - 16t^2 feet at time t seconds.

Suppose you are on top of a 48 foot tall building and toss a ball straight up with an initial velocity of 32 feet per second. (Assume the ball is 48 feet above the ground when it leaves your hand.)

(a) When does the object reach its highest point?

The formula for position is  s(t) = 48 + 32t - 16t^2 feet at time t.
Thus the formula for velocity is  v(t) = s ' (t) = 32 - 32t feet per second at time t.
The object reaches its highest point when its velocity is zero, i.e. when
  v(t) = 0
  32 - 32t = 0
t = 1 second .

(b) When does the object strike ground?
When s(t) = 0
 48 + 32t - 16t^2 = 0
 -16 (t^2 - 2t - 3) = 0
-16 (t - 3) (t + 1) = 0
t = 3 seconds .

(c) What is the object's velocity at the instant it strikes the ground?
v(3) = 32 - 32 (3) = -64feet per second.

(d) When does the object have a velocity of 8 feet per second?
When v(t) = 8
  32 - 32t = 8
  -32t = -24
  t = -24/-32 =3/4 second.


Created by Mathematica  (January 14, 2004)