Download Math 111- Solution of Test 4 Problem 1. Find the antiderivative F of

Math 111- Solution of Test 4
√
x 5
2
Problem 1. Find the antiderivative F of the function: f (x) = 3 + 5x4 +
+ + sin x + √
2
x
1−x
1
1
1
ISolution: Since f (x) = 3 + 5x4 + x 2 + 5 + sin x + 2 √
, then
x
1 − x2
1
x5
x 2 +1
2 3
+ 5 ln x − cos x + 2 arcsin x + c = 3x + x5 + x 2 + 5 ln |x| − cos x + 2 arcsin x + c
F (x) = 3x + 5 + 1
3
5
2 +1
Calculus I - Fall 2003 - Yahdi
Problem 2. Find the absolute extremum values of f (x) = 10 + 27x − x3 on the interval [0, 4].
ISolution: f 0 (x) = 27 − 3x2 = 0 =⇒ 3(9 − x2 ) = 0 =⇒ 9 − x2 = 0 =⇒ x = ±3. Therefore, x = 3 is the only critical
number in the interval [0, 4].
Now, we compare the value of the function at the endpoints and critical numbers in [0, 4]: f (0) = 10, f (3) = 64 &
f (4) = 54.Thus, f has an absolute maximum of 64 at x = 3, and an absolute minimum of 10 at x = 0 .
Problem 3. Below is the graph of the derivative f 0 of a function f , defined on the interval [0, 7].
(1) On what intervals is f increasing? decreasing? Justify your answers to receive credit.
(2) For what values of x does f have a local maximum? local minimum? Justify your answers.
(3) On what intervals is f concave up? concave down? Justify your answers to receive credit.
(4) For what values of x does f have an inflection point? Justify your answers.
ISolution: From the graph of the derivative f 0 , we have the following sign chart of f 0 (positive if above the x-axis and
negative if below the x-axis): 0
⊕
1
3
⊕
7
0
(1) f is increasing on the intervals [0, 1] and [3, 7] because f is positive in these intervals.
f is decreasing on the interval [1, 3] because f 0 is positive in this interval.
(2) The critical numbers are at x = 1, x = 3 and x = 7 because f 0 (x) = 0 since f 0 intersects the x-axis.
• f 0 changes from positive to negative at x = 1. . Therefore, local maximum at x = 1 .
• f 0 changes from negative to positive at x = 3. So local minimum at x = 3
• x = 7 is an endpoint, so local max and min do not apply.
(3) Using the derivative f 0 , f is concave up when f 0 is increasing (f” positive), and concave down when f 0 is
decreasing. So, f is concave up on the interval [2, 6] and f is concave down on the intervals [0, 2]&[6.7] . In
summary we have: 0
C.D.
2
C.U.
6
C.D.
7
(4) Inflection point are points are where f changes concavity. They occur then at : x = 2 and x = 6
Problem 4. A hockey team plays in an arena with a seating capacity of 15,000 spectators. With the
ticket price set at $15, average attendance at a game has been 11,000. A market survey indicates that for
each dollar the ticket price is lowered, average attendance will increase by 1,000. How should the owners
of the team set the price to maximize their revenue from ticket sales? Show your work to receive credit.
ISolution:
Let R= the revenue function = price × quantity
Let n= the number of times the price of the ticket is lowered by $ 1. Then:
(
price
= $15 − n · ($1) = 15 − n dollars,
quantity = number of spectators = 11, 000spectators + n · (1, 000spectators) = 11, 000 + 1, 000n spectators.
Hence
R(n) = (15 − n)(11, 000 + 1, 000n) = 165, 000 + 4, 000n − 1, 000n2 to maximize.
4,000
R0 (n) = 4, 000 − 2, 000n = 0 ⇒ n = 2,000
= 2 is the only critical number.
00
00
R (n) = −2, 000 ⇒ R (2) = −2, 000 < 0. By the second derivative test, R has a local maximum at n = 2, which
is an absolute maximum since it is the only critical number. The best ticket prices to maximize the revenue is then:
$ 15 − (2) = $13 , with 11, 000 + 1, 000(2) = 13, 000 spectators and a revenue of $ R(2) = $169, 000 .
Problem 5. A ball is thrown upward with a speed of 16 ft/s from the edge of a cliff 320 ft above the
ground. With what velocity does it hit the ground?
ISolution:
• a = Acceleration = Gravitational attraction= −32f t/s2 . Take the antiderivative to get the velocity v:
• v(t) = −32t+c. Since v(0) = +16f t/s (positive because upward), then by substitution, −32(0)+c = 16 ⇒ c = 16.
Thus v(t) = −32t + 16 . Then take the antiderivative to get the position s.
2
• s(t) = −32 t2 + 16t + c = −16t2 + 16t + c. Since s(0) = 320f t, then by substitution, −16(0) + 16(0) + c = 320 ⇒
c = 320. Thus
s(t) = −16t2 + 16t + 320 .
• The ball hits the ground when s(t) = 0. So −16t2 + 16t + 320 = 0 ⇔ −16(t2 − t − 20) = 0 ⇔ −16(t − 5)(t + 4) =
0 ⇔ t = 5 or t = −4. Since the event is happening in the future, thus the ball will hit the ground after
t = 5 seconds . The corresponding velocity is then v(5) = −32(5) + 16 = −144f t/s .
1