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Calculus I                                                                 Final Exam                                   December 10, 2002

Name____________________                             R.  Hammack                                        Score ______
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(1) Evaluate the following limits.

(a)    Underscript[lim , x3] (x^2 - 2x )/(x + 1) =(3^2 - 2 (3) )/(3 + 1) = 3/4

(b)    Underscript[lim , x4] (x^2 - 16)/(x - 4) =Underscript[lim , x4] ((x - 4) (x + 4))/(x - 4) = Underscript[lim , x4] (x + 4) = 8

(c)   Underscript[lim , x4] (4 - x)/(2 - x^(1/2)) = Underscript[lim , x4] (4 - x)/(2 - x^(1/2)) (2 + x^(1/2))/(2 + x^(1/2)) = Underscript[ ... im , x4] (4 - x) ( 2 + x^(1/2))/(4 - x) = Underscript[lim , x4] ( 2 + x^(1/2)) = 4

(d)    Underscript[lim , xπ^+] cot(x) =∞

(e)    Underscript[lim , x∞] (6x^5 + x)/(3x^5 - 8) = 6/3 = 2

(f)    Underscript[lim , x2] sin(2x - 4)/(x^2 - 4) = Underscript[lim , x2] sin ( 2 (x - 2) )/((x + 2) (x - 2)) = Underscript[lim , x2] sin ( 2 (x - 2) )/(2 (x - 2)) 2/(x + 2) = (1) 2/(2 + 2) = 1/2


(2) Use the limit definition of the derivative to find the derivative of the function f(x) = (x + 1)^(1/2).

f ' (x) = Underscript[lim , wx] (f(w) - f(x))/(w - x) = Underscript[lim , wx] ... )^(1/2) - (x + 1)^(1/2))/(w - x) ((w + 1)^(1/2) + (x + 1)^(1/2))/((w + 1)^(1/2) + (x + 1)^(1/2)) =

Underscript[lim , wx] (w + 1 - (x + 1))/((w - x) ((w + 1)^(1/2) + (x + 1)^(1/2))) = Un ... 754;x] 1/((w + 1)^(1/2) + (x + 1)^(1/2)) = 1/((x + 1)^(1/2) + (x + 1)^(1/2)) = 1/(2 (x + 1)^(1/2))


(3) Find the following derivatives.

(a)    d/dx[  3x^8 + 2x + 1 ] =24x^7 + 2

(b)    d/dx[  sin(π) + ln(x) ] =1/x

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

(d)    d/dx[   (2x - 1)/(x + 3) ] =(2 (x + 3) - (2x - 1))/(x + 3)^2 = 7/(x + 3)^2

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

(f)   d/dx[   x e^x] =e^x + x e^x

(g)    d/dx[ cos(5x)^(1/2)   ] =(-5sin(5x))/(2cos(5x)^(1/2))

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

(i)    d/dx[   ln(1 - x e^(-x)) ] =-((1) e^(-x) + x e^(-x)(-1))/(1 - x e^(-x)) = (x e^(-x) - e^(-x))/(1 - x e^(-x))

(j)    d/dx[ x^x  ] =x^x(ln(x) + 1)  (work follows)

y = x^x

ln(y) = ln(x^x)

ln(y) = x ln(x)

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

1/ydy/dx = ln(x) + x1/x

dy/dx = y(ln(x) + 1)

dy/dx = x^x(ln(x) + 1)



(k)    d/dx[ e^( sin^(-1)(x))   ] =e^( sin^(-1)(x))/(1 - x^2)^(1/2)
(4) Simplify the following expressions as much as possible.

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

(b)     e^( 3ln(2)) = e^( ln(2^3)) = 8

(c)     FormBox[RowBox[{RowBox[{log_10, (, 0.001, )}], =}], TraditionalForm] -3

(d)     ln(e^(1/2)) =   ln(e^(1/2)) = 1/2

(e)    ln(e)^(1/2) =1^(1/2) = 1

(f)     ln ( sin(π/2) ) =  ln ( 1 ) = 0

(g)     cos^(-1)(1/2) =π/3

(h)    tan ( cos^(-1)(x)) =(1 - x^( 2))^(1/2)/x

(i)    cos(5π) =-1

(j)    sec(-π/4) =1/cos(-π/4) = 1/1/2^(1/2) = 2^(1/2)


(5)   Sketch the graph of y = 1 - e^x

The graph of y = e^xis skecthed in black, that of  y = -e^xis in red.
Finally, the graph of  y = 1 - e^xis done in green.

[Graphics:HTMLFiles/ExF02_67.gif]

(6)   Solve the equation   ln(1/x) + ln(2x^3) = ln(3)   for x.
  
   ln(1/x) + ln(2x^3) = ln(3)  
   ln((2x^3)/x) = ln(3)   
   ln(2x^2) = ln(3)  
  2x^2 = 3
  x^2 = 3/2
  x = (3/2)^(1/2)
  
  Note. The negative root is not a solution because it doesn't check back.

(7) The questions on this page concern the one-to-one function f(x) = x/(2 - x)

(a)    Underscript[lim , x2^+] f(x) =-∞

(b)    Underscript[lim , x∞] f(x) =-1

(c)   List the vertical asymptotes of f  (if any). The line x = 2 is a vertical asymptote.

(d)   List  the horizontal asymptotes of f  (if any). The line y = -1 is a horizontal asymptote.

(e)   Find the inverse of f.

y = x/(2 - x)

y(2 - x) = x

2y - y x = x

y x + x = 2y

x(y + 1) = 2y

x = (2y)/(y + 1)

Now interchange x and y:
y = (2x)/(x + 1)

f^(-1)(x) = (2x)/(x + 1)


(f) State the domain of f.  All values of x except 2.

(g)   State the domain of f^(-1). Looking at the answer from part e above, you can see the domain is all values except -1.

(h)  State the range of  f.  (Range of f) = (Domain of f^(-1)) = all values except -1.

(i)   State the range of f^(-1). (Range of f^(-1)) = (Domain of f) = all values except 2.

(j) Find the equation of the tangent line to the graph of y = f(x) at the point where x = 1.

f ' (x) = ((1) (2 - x) - x(-1))/(2 - x)^2 = 2/(2 - x)^2
Slope of tangent is f ' (1) = 2/(2 - 1)^2 = 2.
Point on tangent is (1, f(1)) = (1, 1/(2 - 1)) = (1, 1).
Using point-slope formula, we get:
y - 1 = 2 (x - 1)
y = 2x - 1

(8) This problem concerns the function f(x) = ln(x^2 + 1).

(a) f ' (x) =(2x)/(x^2 + 1)

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

(c)  Find the interval(s) on which  f  is increasing and on which it is decreasing.

Looking at the form of f ', you can see it's positive if x is positive and negative if x is negative.  Thus f increases on the interval (0, ∞) and decreases on (-∞, 0).

(d)  Find the interval(s) on which  f  is concave up and on which it is concave down.

Looking at the factored version of the second derivative, above, we get:

x -------------------- -1 ---------------  1 ----------------
(1+x)     - - - - - - - - + + + + + + + + + + + + + + +
(1-x)     + + + + + + + + + + + + + +  - - - - - - - - -
(x^2 + 1)^2 + + + + + + + + + + + + + + + + + + + + +
f ''(x)     - - - - - - - - + + + + + + + + - - - - - - - - - -

Conclusion:
f  concave up on  (-1, 1).
f  concave down on  (-∞, -1) and  (1, ∞).

(e)  List the x-coordinates of all inflection points of  f.
-1 and 1

(f)  List all the critical numbers of f.
Only 0

(g)  List the x-coordinates of the relative extrema of  f  (if any) and say whether there is a relative minima or a relative maxima.

Relative minimum at x = 0

(9) Given the equation   y + sin(y) = x,  find  dy/dx.
d/dx[y + sin(y)] = d/dx[x]

dy/dx + cos(y) dy/dx = 1

dy/dx (1 + cos(y) ) = 1

dy/dx = 1/(1 + cos(y))

(10) A rectangular box with with two square sides and an open top is to have a volume of 36 cubic feet. Find the dimensions of the container with minimum surface area.
[Graphics:HTMLFiles/ExF02_118.gif]
Let the dimensions of the box be x by x by y.
Then its volume is x^2y = 36, and we get y = 36/x^2.

Now, the total surface area is
S = 2x^2 + 3x y

S = 2x^2 + 3x 36/x^2

S = 2x^2 + 108/x

So we want to find the minimum of the above function. Setting the derivative equal to 0, we get:

dS/dx = 4x - 108/x^2 = 0

4x = 108/x^2

x^3 = 108/4

x^3 = 27

x = 3

Thus the critical point is 3.
                 3
------------+----------
dS/dx - - - | + + + +

Checking the sign of the derivative, we see that there is a global minimum at x = 3.
For this x value, we have a volume of  3^2y = 36, so y = 4.

Answer: The box should have dimensions of 3 by 3 by 4.

(11) A spherical balloon is deflated in such a way that its radius decreases at a constant rate of 15 cm/min. At what rate is air escaping when the radius is 2 cm?

The volume of the balloon is V = 4/3π r^3.
We know that dr/dt = -15
We want to find dV/dt .

d/dt[V] = d/dt[4/3π r^3]

dV/dt = 4/3π 3 r^2dr/dt

dV/dt = 4/3π 3 r^2 (-15 )

dV/dt = -60π 2^2 = -240π cubic cm. per minute.