Fundamentals of Mathematics
Mathematics 1710 Internet Second Test
February 4, 2006
Instructions: Do all of the following problems. Present your solutions in the accompanying booklet(s). Present your solutions in the order that they appear on this paper. No
books, notes, or crib sheets are allowed. A calculator is allowed. Time allowed - until 1:00
P.M.
1. (a) Given a function f define what it means for f to have an inflection point at a
point.
See definition 7 - page 10 of Calculus III
(b) Describe in detail the 2nd derivative test.
Solution: See Result 5 - bottom of page 7 of Calculus III
(c) State the Fundamental Theorem of Algebra. Solution: See Theorem 8 - page
16 of the Functions document
2. Let f : [a, b] → R be a function.
(a) What does it mean for f to have a global maximum at a point c, for a ≤ c ≤ b?
Solution: See Definition 5 p. 8 of Calculus II
(b) What does it mean for f to have a local minimum at a point c, for a < c < b?
Solution: See Definition 6 p. 8 of Calculus II
3. Given the function f (x) = x2 −5x+7 complete the square on the expression x2 −5x+7.
Using the result, state at what point c the function takes on its minimum value.
Finally what is the minimum functional value?
Solution: Completing the square gives: f (x) = (x − 52 )2 − 25
4 + 7. Thus the minimum
3
value occurs at x = 52 and f ( 52 = − 25
+
7
=
.
4
4
4. Let y = x3 and suppose the graph is moved 1 unit to the left, one unit down, reflected
about the x axis and flattened by a factor of 1/2 . What is the equation that describes
the new graph?
Solution:
• moving to the left gives: y = (x + 1)3
1
• moving one unit down gives: y = (x + 1)3 − 1
• reflecting about x axis gives: y = −(x + 1)3 + 1
• flatening by
1
2
gives: y = − 12 (x + 13 +
1
2
5. Calculate the derivative of the following functions
4
2
(a) f (x) = x 3 − x 3
1
1
Solution f 0 (x) = 43 x 3 − 23 x− 3
(b) G(y) = (y 2 + 1)(2y − 7)
Solution: G0 (y) = 2y(2y − 7) + (y 2 + 1)2
6. Calculate the derivative of the following functions
4x3
(a) h(x) = √
2−x
Solution:
h0 (x)
√
1
12x2 ( 2 − x) − 4x3 (2 − x)− 2 (−1)
=
2−x
(b) g(x) = (x5 + 3)99
Solution: g 0 (x) = 99(x5 + 3)98 5x4
√
7. Find the equation of the tangent line to the graph of the function f (x) =
the point (4, 25 ).
Solution: Calculating the derivative,
0
f (x) =
1 − 12
(x
2x
1
And
1
+ 1) − x 2
(x + 1)2
which becomes
f 0 (x) =
x
at
x+1
1
2x 2
1
(x + 1) − x 2
(x + 1)2
1
4 (5)
−2
3
=−
25
100
which is then the slope of the tangent line. Knowing that the tangent line must pass
through the point (4, 52 ), in the general equation of a line y = mx + b we must now
3
have 52 = − 100
4 + b. Solving for b gives, b = 13
25 . Thus the equation of the tangent line
is
3
13
x+
y=−
100
25
f 0 (4) =
2
8. At noon a ship A is 150 km west of ship B. Ship A is sailing east at 35 km/hr and
ship B is sailing north at 25 km/hr. How fast is the distance between the ships
changing at 4:00 PM?
bf Solution: The trick is to have the right diagram, and for this it was necessary to
have understood what the problem was asking.
Figure 1: Ship A traveling east at 35 km/hr and ship B traveling north at 25 km/hr after
t hours
Here (d(t))2 = (150 − 35t)2 + (25t)2 so that differentiating both sides using the chain
rule gives
2d(t)d0 (t) = −2(150 − 35t)35 + 2(25t)25
so that
d0 (t) =
We need to find d0 (4). Since d(4) =
1850t − 5250
d(t)
√
10, 100, we have
d0 (4) = √
2150
≈ 21.39km/hr
10, 100
9. A paper cup has the shape of a cone with height 10 cm and radius 3 cm (at the top).
If water is poured into the cup at the rate of 2 cm3 / sec, how fast is the water level
rising when the water is 5 cm deep?
Solution: This problem is identical to example 3 on page 5 of the Calculus II document. The only difference is that the numbers have been changed. For a solution
8
.
proceed as in the example. The final answer is h0 (t) = 9π
3
10. Using the concavity test and the second derivative test, for the function f (x) =
4x5 − 3x3 + 1 find the local maximums and minimums and the intervals over which
the graph is concave down and the intervals where the graph is concave up.
Solution: Calculating derivatives we have:
• f 0 (x) = 20x4 − 9x2 = x2 (20x2 − 9)
• f 00 (x) = 80x3 − 18x = 80x(x2 −
18
80 )
q
q
9
3
Setting f
= 0 to find critical points gives: x = 0 or x = ± 20 = ± 2 15 =
√
√
√
3
3
3
5. Using the second derivative test, since f 00 ( 10
5) > 0 and f 00 (− 10
5) < 0,
± 10
√
√
3
3
it follows that there is a relative max at x = − 10 5 and a relative min at x = + 10
5.
But f 00 (0) = 0 so the second derivative test gives no information about the critical
point x = 0.
0 (x)
In order to find areas of concavity, we begin by finding the zeros of f 00 (x)q= 2x(40x2 −
√3
9). Setting this equal to zero and solving for x, gives x = 0 or x = ± 18
80 = ± 40 .
Sign analysis as in the table below, tells us that the graph is concave down the
intervals (−∞, − √340 ) and (0, √340 ) and concave up on the intervals (− √340 , 0) and
(− √340 , ∞).
Figure 2: sign analysis for f 00 (x) = 80x(x2 −
4
18
80 )