Test #1 Math 220 Fall 2014
There are a total of 100 points.
Name:
1. (3 points each) Find the domain of each function (show some work).
a) f (x) =
3
5x + 7
7
Solution: Set denominator equal to 0. So 5x + 7 = 0 means that x = − . Thus our domain
5
is everything but this point. So that Df = (−∞, −7/5) ∪ (−7/5, ∞).
b) g(x) =
x100 + 12
x−3
Solution: Set denominator equal to 0. We get x = 3. So Df = (−∞, 3) ∪ (3, ∞)
c) h(x) = ln(3x − 7).
Solution: Logarithms can only have positive arguments. We set 3x − 7 > 0 to obtain 3x > 7
7
and x > . So the domain is Df = (7/3, ∞).
3
1
d) j(x) = √
.
2x + 1
Solution: We cannot divide by zero, nor can we take the square root of a negative number.
1
So we need 2x + 1 > 0, which means x > − . So Df = (−1/2, ∞).
2
√
e) q(x) = e
5x−10
Solution: No negatives under the square root! We need 5x − 10 ≥ 0 and so x ≥ 2. Thus
Df = [2, ∞).
2. (3 pts each) Solve each equation for x (show some work).
a) ex
2 +4x
= e−4
Solution: The bases are equal, so the exponents are equal. x2 + 4x = −4 which means
x2 + 4x + 4 = 0 which factors as (x + 2)(x + 2) = 0, and so x = −2.
b) 52x+1 = 25
Rewrite as 52x+1 = 52 . Now the bases are equal, so the exponents are as well. 2x + 1 = 2
means that x = 1/2.
c) 3x = 6(4x )
!x
3
Solution: Divide both sides by 4x to get
= 6. Take your favorite log of both sides.
4
!x
!
3
3
= ln(6). We can pull the x out now, so x · ln
= ln(6).
I will use ln. So that ln
4
4
ln(6)
!.
And finally, x =
3
ln
4
d) ln(5x − 2) = 9
Solution: Put both sides ontop of an e to cancel out the ln. We get eln(5x−2) = e9 so that
e9 + 2
9
.
5x − 2 = e . Thus, x =
5
e) log6 (x + 1) − log6 (x + 3) = log6 (3)
Solution: We must combine the logs into one like this log6
x+1
= log6 (3). The log bases
x+3
x+1
= 3 which means x + 1 = 3(x + 3)
x+3
and so x + 1 = 3x + 9. Finally, x = −4. If you plug x = −4 into the original equation, you
get the log of a negative number. Thus there is no solution.
are equal so we know the arguments are equal. So
3. (3 points) Nigel invests $3000 at 10% compounded continuously for 10 years. Will he
have more or less than $9,000 in his account at the end of this time? Hint: e ≈ 2.72.
Solution: Use F = P ert . Here, F = 3000e(.1)(10) = 3000e1 = 3000e. Use your approximation
skills to deduce 3000e < 9000. So he will have less than 9000 dollars.
4. (5 points) Suppose you want to buy a fancy sandwich that costs $100. You have only
$50, and you decide to invest the money in the bank until you can buy the sandwich. Your
bank offers an interest rate of 8% compounded continuously. How long must you wait to
purchase your sandwich? Do NOT simplify your answer.
Solution: Since the interest is continuous, use F = P ert . Here we will need to solve for t.
So 100 = 50e(.08)(t) . Thus, 2 = e.08t . We can take ln of both sides to obtain ln(2) = .08t. So,
ln(2)
t=
.
.08
5. (5 points) Ash and Brock would like to have $100,000 in their bank account 20 years
from now. Their bank offers an investment rate of 4% compounded semi-annually. How
much money should they invest to achieve their goal? Do NOT simplify your answer!
r
Solution: We use P = F (1 + )−nt since we are looking for the principal. Notice n = 2
n
because we compound semi-annually. Substituting everything in, we have
P = 100000(1 +
.04 (−2)(20)
.04 −40
)
= 100000(1 +
)
2
2
No further simplification is necessary here.
6. (25 points) Find the following limits, if they exist (show some work). If the limit doesn’t
exist, say why.
x2 + 1
a) lim+
x→5 x − 5
The numerator goes to 52 + 1 = 26 > 0 as x → 5+ . The denominator goes to 0+ as x → 5+ .
x2 + 1
Thus, lim+
= +∞.
x→5 x − 5
9−x
√
x→9 3 −
x
b) lim
We must multiply by the conjugate of the denominator on the top and bottom. We get
√
√
√
9−x
9 − x (3 + x)
(9 − x)(3 + x)
√ = lim
√
√ = lim
lim
= lim (3 + x) = 6
x→9 3 −
x→9
9−x
x x→9 3 − x (3 + x) x→9
x2 − 3x − 10
x→5
x+2
c) lim
Solution: We cannot evaulate directly. Factor the numerator to obtain:
x2 − 3x − 10
(x − 5)(x + 2)
= lim
= lim (x − 5) = 0
x→5
x→5
x→5
x+2
x+2
lim
d) lim ex
2 +5x−1
x→∞
Solution: x2 + 5x − 1 goes to ∞ as x → ∞. Thus, lim ex
x→∞
2 +5x−1
=∞
e) lim log(x + 1) − log(x4 )
x→∞
Solution: First, combine it into one piece using the subtraction rule of logs.
lim log(x + 1) − log(x4 ) = lim log
x→∞
x→∞
x+1
x4
Now notice that the argument of the log goes to 0 as x goes to infinity. Then remember that
as the argument of the log function goes towards 0, the function goes off to negative infinity.
So the answer is −∞.
7. (20 points) For each of the points listed, answer the following questions: Is the point in
the domain of f (x)? What are the one-sided limits? What is the two-sided limit? Is f (x)
continuous at this point (use the formal definition for full credit)?

x+4


 2
x − 16
f (x) =
x+1



5
x < −4
−4 ≤ x ≤ 0
0≤x<6
x>6
a. x = 0
This point is in the domain since 0 is listed in the definition of f (x).
lim f (x) = 1 and also lim− f (x) = −16.
x→0+
x→0
The 2-sided limit does not exist and therefore f (x) is not continuous at x = 0.
b. x = 1
This point is contained in the interval [0, 6), so it is in the domain.
lim f (x) = 2 = lim− f (x), so the 2-sided limit exists and are equal to 2.
x→1+
x→1
Also, lim f (1 + h) = lim (1 + h) + 1 = 2 = f (1), so f is continuous at x = 1.
h→0
c. x = 6
h→0
This point is not in the domain (look at the intervals closely). Also, lim+ f (x) = 5 and
x→6
lim− f (x) = 7. So the 2-sided limit does not exist. Thus the function is not continuous at
x→6
this point.
d. x = −4
This point is in the domain. Additionally, lim+ f (x) = 0 and lim− f (x) = 0. Thus the
x→4
x→4
2-sided limit exists and is equal to 0.
Moreover, lim f (−4 + h) = lim (−4 + h)2 − 16 = 0 = f (−4), so our function is continuous
h→0
h→0
here.
8. (12 points) A factory is manufacturing rectangular wooden doors that come with a
number of circular glass windows, and a number of plastic pieces used to hold the windows
in place. The customer cuts the holes out for the windows by themself, so we are concerned
only with the production of the rectangular door, the circular windows, and the plastic
pieces. The cost of the lumber to create one of the doors is $1.50 per square inch. The cost
of the glass to make the windows is 50 cents per square inch. The cost of a plastic piece is
25 cents.
Write a function of the cost in dollars to produce a single door which depends on the
following four variables:
r = radius of the windows
w = width of the door
l = length of the door
N = number of windows
You may keep π as a symbolic value – no need to approximate.
Solution: The formula is
C(r, w, l, N ) = 1.50(l)(w) + N (.5) · πr2 + N (.25)
**I also gave full credit if you wrote
C(r, w, l, N ) = 1.50((l)(w) − πr2 · N ) + N (.5) · πr2 + N (.25)
This formula takes into account the cost of the door if the cut-out holes for the windows
don’t factor into the cost of the wood. This is technically wrong since it is not what the
question asked for, but I gave credit anyway because it is a harder problem to solve.