### Solutions! - Stony Brook Mathematics

```Quiz 1
Problem 1. The slope is given by the slope formula, so
−1 − 1
= −1/3.
m=
4 − (−2)
Using the point-slope formula for a line, we get the equation
y = − 13 (x − 4) − 1.
Problem 2. We need to find when the revenue function is equal to the cost function.
This gives the equation
10q + 500 = 30q.
Solving, we find that
q = 25.
Problem 3. The slope formula gives
12, 000, 000 − 60, 000, 000
= −3, 000, 000.
1986 − 1970
This is in units of typewriters per year. The number means that, on average, US households had 3 million fewer typewriters every year.
1
Quiz 2
Problem 1. The initial value is 100 and the annual rate of increase was 1.15. Thus
there are
P (t) = 100(1.15)t
tigers after t years.
Problem 2. Taking the logarithm of both sides, and then applying the rules of
logarithms,
ln(4e2t ) = ln(10)
ln(4) + ln(e2t ) = ln(10)
ln(4) + 2 ln(et ) = ln(10)
ln(4) + 2t = ln(10)
2t = ln(10) − ln(4)
ln(10/4)
2
ln(5/2)
t=
2
t=
Note that
ln(4e2t ) 6= 2 ln(4et ).
Try plugging in random values for t on a calculator if you don’t believe this.
2
Quiz 3
Problem 1.
f (g(x)) = (g(x))2 − 2g(x) − 8
= (x + 3)2 − 2(x + 3) − 8
= (x2 + 6x + 9) − 2x − 6 − 8
= x2 + 4x − 5
Problem 2. Since y(0) is the initial value, C = 5. This means that y(x) = 5eαx .
Using the other given piece of information,
1 = y(1) = 5eα .
Solving the equation, one finds
α = ln(1/5).
So
y(x) = 5eln(1/5)x
and
−1 )
y(−1) = 5e− ln(1/5) = 5eln((1/5)
3
= 5eln(5) = 25.
Quiz 4
Problem 1. The formula for average velocity gives
3(2 + h)2 + 5 − 3 · 22 − 5
12 + 12h + 3h2 − 12
s(2 + h) − s(2)
=
=
= 12 + 3h
2+h−2
h
h
Problem 2.
2 · (3 + h)2 − 2 · 32
18 + 12h + 2h2 − 18
lim
= lim
h→0
h→0
h
h
2
12h + 2h
= lim
h→0
h
h(12 + 2h)
= lim
h→0
h
= lim 12 + 2h
h→0
= 12
4
Quiz 5
Problem 1.
(x + h)2 + 1 − x2 − 1
h→0
h
2
x + 2xh + h2 − x2
= lim
h→0
h
2xh + h2
= lim
h→0
h
h(2x + h)
= lim
h→0
h
= lim 2x + h
f 0 (x) = lim
h→0
= 2x
Problem 2. We could use the limit definition,
we will make calculation errors if we use the rules
deratives. Thus
d
d
y=
(3x3 − 2x − 8)
dx
dx
d
d
d
=
(3x3 ) + (−2x) + (−8)
dx
dx
dx
d
d
= 3 (x3 ) − 2 (x)
dx
dx
= 9x2 − 2
5
but it is quicker and less likely that
that we have learned for calculating
by sum law
because constants cross
d
dx
by derivatives of power laws
Quiz 6
Problem 1. Recall that the formula for the tangent line to f (t) at t = a is given by
the formula
y = f 0 (a)(t − a) + f (a).
In this case, f (t) = y(t), and we may calculate that
y 0 (t) = 3e3t
so that
y 0 (0) = 3.
Thus the equation for the tangent line is given by
y = 3t + 4.
Problem 2. There is a slight notational error in the question. We should compute
y 0 (t) and y 00 (t). This results in
y 0 (t) = 5/x + 7et − 8t.
Taking the derivative again,
y 00 (t) = −5/x2 + 7et − 8.
6
Quiz 7
Problem 1.
(1) We use the rules of differentiation:
d
d
d t
(e + 3t) = (et ) + (3t)
dt
dt
dt
= et + 3.
(2) Now we must use the chain rule. The inner function is ex + 3x and the outer
function is ln(x).
d
1
(ln(ex + 3x)) = x
· (ex + 3)
dx
e + 3x
Problem 2.
(1) Now we use the product rule.
d −2 3x
(x e ) = (−2x−3 )(e3x ) + (x−2 )(3e3x )
dx
(2) Plugging x = 1 into the previous answer, we get
h0 (1) = (−2(1)−3 )(e3(1) ) + ((1)−2 )(3e3(1) ) = −2e3 + 3e3 = e3 .
7
Quiz 8
Problem 1. First we need to find the critical points of f (x)—those values of x = a
for which f 0 (a) = 0. So we must solve the equation
f 0 (x) = 3x2 − 4x = 0.
We can solve this by factoring:
0 = 3x2 − 4x = x(3x − 4).
The solutions are thus x = 0 and x = 4/3.
We use the second derivative test to determine if these are local minima or maxima.
First note that the second derivative of f (x) is
f 00 (x) = 6x − 4.
Thus f 00 (0) = −4 and so 0 is a maximum.
On the other hand, f 00 (4/3) = 4, so 4/3 is a minumum.
Problem 2. By the rules of integrals,
Z
Z
Z
Z
5
5
(x + x − 1) dx = x dx + x dx − 1 dx
= x6 /6
+ x2 /2
8
−x
+ c.
```