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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)

(b)

(c)

(d)

(e)

(f)

(2) Use the limit definition of the derivative to find the derivative of the function .

(3) Find the following derivatives.

(a)

(b)    1/x

(c)

(d)

(e)

(f)

(g)

(h)

(i)

(j)      (work follows)

(k)
(4) Simplify the following expressions as much as possible.

(a)

(b)

(c)      -3

(d)       1/2

(e)

(f)

(g)

(h)

(i)    -1

(j)

(5)   Sketch the graph of

The graph of is skecthed in black, that of  is in red.
Finally, the graph of  is done in green.

(6)   Solve the equation      for x.

Note. The negative root is not a solution because it doesn't check back.

(a)

(b)

(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.

Now interchange x and y:

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

(g)   State the domain of. 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 ) = all values except -1.

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

(j) Find the equation of the tangent line to the graph of at the point where .

Slope of tangent is .
Point on tangent is .
Using point-slope formula, we get:

(8) This problem concerns the function .

(a)

(b) =

(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 and decreases on .

(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)     + + + + + + + + + + + + + +  - - - - - - - - -
+ + + + + + + + + + + + + + + + + + + + +
f ''(x)     - - - - - - - - + + + + + + + + - - - - - - - - - -

Conclusion:
f  concave up on  .
f  concave down on   and  .

(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   ,  find  .

(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.

Let the dimensions of the box be x by x by y.
Then its volume is , and we get .

Now, the total surface area is

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

Thus the critical point is 3.
3
------------+----------
- - - | + + + +

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  , 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 .
We know that dr/dt = -15
We want to find dV/dt .

cubic cm. per minute.