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MATH 148.05A
Final Exam
March 23, 2011
Name
1
2
3
4
5
6
7
8
9
Total
•
•
•
•
10
10
10
10
10
10
10
10
10
90
Please check that your exam contains 9 problems on 9 pages, and one page of formulae.
You may use a scientific calculator. No graphing calculators, no smartphone apps.
Please turn all electronics OFF and put them away for the duration of the exam.
Unless otherwise indicated, you must show your work. The correct answer with no supporting work may result in no credit.
• If you use a guess-and-check method when an algebraic method is available, you may not
receive full credit.
GOOD LUCK!
η(q) =
p(q)
q
dp
dq (q)
η(p) =
p
dq
· (p)
q(p) dp
dr
= q(p) · (1 + η(p))
dp
dr
= p(q) · (1 +
dq
xn+1 = xn −
1
η(q) )
f (xn )
f 0 (xn )
Math 148.05A
1
(1) Let
−2
+ 3.
x+1
(a) Find limx→∞ f (x). Show enough work to demonstrate why your answer is correct.
f (x) =
We have
−2
= lim
+ lim 3
x→∞ x + 1
x→∞
x→∞
−2
= lim
+ 3.
x→∞ x + 1
But in this limit, the numerator is -2 and the denominator gets huge. So the fraction
approaches 0. So the limit is 3.
lim
−2
+3
x+1
(b) Find limx→∞ f 0 (x). Show enough work to demonstrate why your answer is correct.
First we need to calculate the derivative of f :
2
d
(−2(x + 1)−1 + 3) = 2(x + 1)−2 =
.
f 0 (x) =
dx
(x + 1)2
Now when we take the limit as x → ∞, again the numerator is a constant (2 in this
case) and the denominator gets huge. So the fraction goes to 0 and the limit is 0.
2
(2) Below is the graph of a function g(x).
(a) Where is g(x) continuous and discontinuous? Briefly explain why your answer is
correct.
The function is discontinuous at −1 and 2, because it jumps at those points. It is
continuous everywhere else.
(b) Where is g(x) differentiable and nondifferentiable? Briefly explain why your answer is
correct.
The function is nondifferentiable at −1 and 2 because it is not continuous there. It
is not differentiable at 1 because the graph is pointy there, so there is no tangent
line. It is also not differentiable at −1/2, because the tangent line is vertical. It is
differentiable everywhere else.
Math 148.05A
3
(3) Estimate f 0 (x) at the points indicated below using the graph of f (x). Draw any points or
lines that you use on the graph.
(a) f 0 (−3)
Since f 0 (−3) is the slope of the tangent line at x = −3, we can draw the tangent line
there on the graph, then calculate its slope (by finding the rise over the run between
two points on that line). I get about f 0 (−3) = 1.
(b) f 0 (1/2)
Again, think about the tangent line at x = 1/2. It looks like the tangent line there
will be horizontal. So f 0 (1/2) = 0.
4
(4) The price-demand equation for a product is
(p + 2)(q + 1) = 20
Find the point eleasticity for p = 2 and p = 12. For each price, is demand elastic or
inelastic?
Our formulae for η assume we have p in terms of q, or q in terms of p. Since we want to
know η for certain values of p, I chose to solve for q in terms of p to get
20
q=
− 1.
p+2
Now calculating
dq
dp
gives
dq
−20
=
.
dp
(p + 2)2
Now we’re ready to calculate η:
p dq
·
q dp
p
−20
= 20
·
(p + 2)2
p+2 − 1
η=
=
−20p
.
20(p + 2) − (p + 2)2
Now I can plug in the values for p that I want. η(2) = −0.625, so since | − 0.625| < 1
demand at this price is inelastic. On the other hand, η(12) = −2.86, so since | − 2.86| > 1,
demand at price p = 12 is elastic.
Math 148.05A
5
(5) Calculate
Z
Z
(x + 1)(ex
2 +2x
− x2 ) dx.
2
(x + 1)(ex +2x − x2 ) dx
Z
Z
x2 +2x
= (x + 1)e
dx − (x + 1)x2 dx
At this point, I’ll do a u-substitution on the left integral, using u = x2 + 2x. Then
du = (2x + 2)dx, and so 12 dy = (x + 1)dx. So, continuing the calculation, I get
Z
2
(x + 1)(ex +2x − x2 ) dx
Z
Z
1 u
=
e
dy
−
(x3 + x2 )dx
2
= 21 eu − ( 14 x4 + 31 x3 ) + C
1 2
= ex +2x − 21 x4 − 31 x3 + C
2
6
√
(6) Marginal cost for a product is given by M C(q) = q + 1.
(a) Estimate (using differentials) the change in total cost if q changes from 11 to 10.
We have ∆C ≈ dC = dC
dq dq = M C · dq. But we’re starting at q = 11, and q is changing
by −1, so
√
√
∆C ≈ M C(11) · dq = 12 · (−1) = − 12.
√
So when we product 10 units instead of 11, costs go down by 12.
(b) Find the exact change in cost if q changes from 11 to 12. (Hint: ∆y =
Rb
a
y 0 (x)dx.)
We have
Z
12
∆C =
11
12
dC
dq
dq
Z
M C(q)dq
=
=
11
Z 12 p
q + 1dq
12
= 23 (q + 1)3/2 11
11
= 32 (133/2 − 123/2 )
≈ 3.535.
(c) If fixed costs change, briefly describle how this affects the previous two calculations.
Neither part (a) nor (b) depended on fixed costs, only on marginal cost. So a change
in fixed costs would not have changed either calculation.
Math 148.05A
(7) The rate of change of price with respect to time is
dp
1
= + t2/3 + t4/3 .
dt
t
The price at t = 2 is 10. Find the price function p(t).
We have
Z
dp
dt
dt
Z 1
2/3
4/3
dt
=
+t +t
t
p(t) =
= ln |t| + 35 t5/3 + 37 t7/3 + C.
Now we can use the initial condition p(2) = 10:
10 = p(2) = ln 2 + 35 25/3 + 37 27/3 + C.
Solving for C gives C ≈ 5.242. So price as a function of time is
ln |t| + 53 t5/3 + 37 t7/3 + 5.242.
7
8
(8) Below is a graph of f (x). Draw a picture of how you would estimate the area under f (x)
between x = 1 and x = 6 using four rectangles.
You would split the interval from x = 1 to x = 6 up into 4 pieces. On each piece, you
would draw a rectangle based on that subinterval and whos height is given by the function
at the left (or right) hand endpoint of that subinterval.
Math 148.05A
9
(9) Find the area of the region pictured below:
First we must find the points where the two curves intersect. We can solve both equations
for x and then set them equal:
x = y 2 − y,
x = y + 3.
Setting them equal and solving gives y = −1 and y = 3, just like the picture shows. Since
we have x in terms of y, we’ll break the region up into horizontal strips. Then the area is
the integral of the curve on the right minus the curve on the left. So we get
Z 3
area =
((y + 3) − (y 2 − y))dy
−1
3
Z
=
(−y 2 + 2y + 3)dy
−1
3
= − 31 y 3 + y 2 + 3y −1
= 10 23 .