MATH 31B: MIDTERM 1 REVIEW
JOE HUGHES
1. Inverses
1. Let f (x) =
1
x−3 .
Find the inverse g(x) for f .
Solution: Setting y = (x − 3)−1 and solving for x gives
yx − 3y = 1
and
x=
Therefore g(x) =
1 + 3y
y
1+3x
x .
2. Let f (x) = x4 + 32x. Find a domain on which f is invertible. If g is the inverse of f ,
determine g 0 (33).
Solution: f 0 (x) = 4x3 + 32, which is non-negative when x ≥ −2. Therefore f is increasing,
hence one-to-one, on [−2, ∞).
After checking f (x) for some small values of x, we find that
f (1) = 1 + 32 = 33
so g(33) = 1. By the inverse function theorem,
g 0 (33) =
1
1
1
1
= 0
=
=
f 0 (g(33))
f (1)
4(1) + 32
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2. Compound interest and present value
A new computer system costing 25,000 dollars will reduce labor costs by 7000 dollars per
year for 5 years. If r = 0.08, is this a good investment?
Solution: The present value of the savings is
7000(e−.08 + e−.16 + e−.24 + e−.32 + e−.4 ) ≈ 27, 708.5
so this is a good investment.
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JOE HUGHES
3. y 0 = k(y − b)
A cold metal bar at −30 degrees Celsius is submerged in a pool maintained at a temperature
of 40 degrees Celsius. Half a minute later, the temperature of the bar is 20 degrees Celsius.
How long will it take the bar to attain a temperature of 30 degrees Celsius?
Solution: The differential equation for the law of cooling is y 0 = −k(y − T0 ), where T0 is
the ambient temperature. The solution to this differential equation is
y(t) = T0 + Ce−kt
where C = y(0) − T0 . In our case, y(0) = −30 and T0 = 40, so C = −70. To determine k,
we plug in t = 21 :
k
1
20 = y( ) = 40 − 70e− 2
2
or
k
70e− 2 = 20
Dividing by 70 and taking logarithms gives
k = −2 ln(2/7) = 2 ln(7/2)
where we have used the identity − ln x = ln(1/x).
To determine the time when the bar reaches 30 degrees, set y(t) = 30 and solve for t:
30 = y(t) = T0 + Ce−kt = 40 − 70e−2 ln(7/2)t
or
e−2 ln(7/2)t =
1
7
Taking logarithms gives
−2 ln(7/2)t = ln(1/7)
so that
t=
ln(7)
− ln(1/7)
=
≈ 0.7766
2 ln(7/2)
2 ln(7/2)
(in units of minutes).
4. L’Hopital’s Rule
1. Evaluate limx→0
sin 4x
.
x2 +3x+1
Solution: Before using L’Hopital’s rule, always check that the limit is in fact an indeterminate form. In this case, taking f (x) = sin 4x and g(x) = x2 + 3x + 1, we see that
lim f (x) = 0
x→0
but
lim g(x) = 1
x→0
MATH 31B: MIDTERM 1 REVIEW
3
Therefore L’Hopital’s rule does not apply, but we can simply plug in 0 to obtain
sin 4x
=0
lim
x→0 x2 + 3x + 1
2. Evaluate limx→ π2 (x − π2 ) tan x.
sin x
cos x
Solution: Writing tan x =
gives
limπ (x −
x→ 2
(x − π2 ) sin x
π
) tan x = limπ
2
cos x
x→ 2
and g(x) = cos x. Then f ( π2 ) = 0 = g( π2 ) and
π
g 0 (x) = − sin x
f 0 (x) = sin x + (x − ) cos x
2
which are both non-zero near π2 . Therefore L’Hopital’s rule can be applied, yielding
Let f (x) = (x −
π
2 ) sin x
limπ
x→ 2
(x − π2 ) sin x
sin x + (x − π2 ) cos x
= limπ
= −1
cos x
− sin x
x→ 2
3. Evaluate limx→0+ xsin x .
Solution: Let L = limx→0+ xsin x . Then
ln(L) = lim sin x ln(x) = lim
x→0+
x→0+
ln(x)
1
sin x
Let f (x) = ln(x) and g(x) = (sin x)−1 . Then
lim ln(x) = −∞
x→0+
lim (sin x)−1 = ∞
x→0+
and
cos x
1
g 0 (x) = − 2
x
sin x
which are both non-zero near 0 (though not defined at 0). Therefore L’Hopital’s rule
gives
1
ln(x)
sin2 x
x
lim
=
lim
=
lim
−
− cos x
1
x cos x
x→0+
x→0+
x→0+
sin x
sin2 x
f 0 (x) =
Now let f (x) = sin2 x and g(x) = x cos x. Both f and g are zero at x = 0, and
f 0 (x) = 2 sin x cos x
g 0 (x) = cos x − x sin x
g 0 (x) is non-zero near 0, and f 0 (x) is non-zero for all x 6= 0 that are sufficiently close to
0.
Therefore L’Hopital’s rule can be applied again to give
lim −
x→0+
sin2 x
2 sin x cos x
= lim −
=0
x cos x x→0+ cos x − x sin x
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JOE HUGHES
Thus ln(L) = 0, so L = e0 = 1.
5. Integration by Parts
1. Evaluate
R
2
x3 ex dx.
Solution: First make the substitution u = x2 . Then du = 2xdx, hence
Z
Z
Z
1
1
2 x2
3 x2
x e 2x dx =
ueu du
x e dx =
2
2
Now integrate by parts to get
Z
ueu du = ueu − eu + C
and substitute the definition of u in to obtain
Z
i
1h
2
2
2
x3 ex dx = x2 ex − ex + C
2
R
2. Evaluate ln(x)2 dx.
Solution: Integrate by parts with u(x) = ln(x)2 and v 0 (x) = 1. Then
Z
Z
Z
2 ln(x)
ln(x)2 dx = x ln(x)2 − x
dx = x ln(x)2 − 2 ln x dx
x
Integrate by parts a second time, now with u(x) = ln x and v 0 (x) = 1, to obtain
Z
ln x dx = x ln x − x + C
Therefore
Z
R√
ln(x)2 dx = x ln(x)2 − 2x ln x + 2x + C
√
xe x dx.
√
√ , hence
Solution: Let u = x. Then du = 2dx
x
Z
Z
Z
√
√ dx
√
x
x
√ = 2 u2 eu du
xe dx = 2 xe
2 x
Now integrate by parts to get
Z
u2 eu du = u2 eu − 2ueu + 2eu + C
3. Evaluate
and plug in the definition of u to obtain
Z
√
√
√ √
√ √x
xe dx = 2xe x − 4 xe x + 4e x + C