MATH 1101, solutions to practice problems for the final exam
1. Compute the limit:
a) limx→3 e1/x
Answer: e1/3 .
b) limx→3 ln(x − 3)
Answer: −∞.
c) limx→3
ln(x−2)
x−3
Solution: As x approaches 3, (x−2) approaches 1, so ln(x−2) approaches
ln(1) = 0. Therefore we have a limit of the form 0/0 and can apply the
L’Hôpital’s rule:
lim
x→3
d) limx→∞
ln(x − 2)
1/(x − 2)
= lim
= 1.
x→3
x−3
1
8x5 −7x3 +9
(3x2 −1)(2x3 −3)
Solution: Let us divide the top and the bottom of this fraction by x5 :
8x5 − 7x3 + 9
8 − 7/x2 + 9/x5
8−0+0
4
=
lim
=
=
.
x→∞ (3x2 − 1)(2x3 − 3)
x→∞ (3 − 1/x2 )(2 − 3/x3 )
(3 − 0)(2 − 0)
3
lim
e) limx→∞
ex
x3
Solution: As this is the limit of the type ∞/∞, we can apply the
L’Hôpital’s rule several times:
ex
ex
ex
ex
=
lim
=
lim
=
lim
= ∞.
x→∞ x3
x→∞ 3x2
x→∞ 6x
x→∞ 6
lim
f) limx→0
arctan(x)−x
x3
Solution: As this is the limit of the type 0/0, we can apply the L’Hôpital’s
rule several times:
arctan(x) − x
1/(1 + x2 ) − 1
1 − 1 − x2
=
lim
=
lim
=
x→0
x→0
x→0 3x2 (1 + x2 )
x3
3x2
lim
−x2
−1
−1
= lim
=
.
2
2
2
x→0 3x (1 + x )
x→0 3(1 + x )
3
lim
1
2. Compute the derivative of the following functions:
a) f (x) = x ln x − x
Answer: ln(x)
b) f (x) = e3x
2
2
Answer: 6xe3x .
c) f (x) = (x − 1)5 cos x
Answer: 5(x − 1)4 cos x − (x − 1)5 sin x.
d) f (x) = sin(ln x)
Answer:
cos(ln x)
.
x
e) f (x) =
e3x+2
cos x+2
Answer:
(3 cos x+6+sin x)e3x+2
.
(cos x+2)2
3. Find the minimal and maximal values of a function:
a) f (x) = x2 e−x on [0, 1]
Solution: We have f 0 (x) = 2xe−x − x2 e−x = x(2 − x)e−x , so f 0 (x) = 0
for x = 0 and x = 2. Since x = 2 is not on the interval and x = 0 is one of
the endpoints, we just need to compute f (0) = 0 with f (1) = e−1 . Therefore
the minimal value is 0 and the maximal value is e−1 .
b) f (x) = 2x3 − 3x2 + 1 on [−1, 2]
Solution: We have f 0 (x) = 6x2 − 6x = 6x(x − 1), so f 0 (x) = 0 for x = 0
and x = 1. We need to compute
f (−1) = −2−3+1 = −4, f (0) = 1, f (1) = 2−3+1 = 0, f (2) = 16−12+1 = 5.
Therefore the minimal value is (−4) and the maximal value is 5.
c) f (x) = sin2 x on [0, π]
Solution: We have f 0 (x) = 2 sin x cos x, so f 0 (x) = 0 when either sin x =
0 (so x = πk) or cos x = 0 (so x = π/2 + πk). On the interval [0, π] we have
3 critical numbers 0, π/2, π and both endpoints are among them. Therefore
one needs to compute f (0) = 0, f (π/2) = 1, f (π) = 0, and the minimal and
maximal values are 0 and 1 respectively.
d)
x
1+x2
on [−2, 2]
2
2
−x(2x)
Solution: We have f 0 (x) = 1+x
= 1 − x2 (1 + x2 )2 , and the critical
(1+x2 )2
numbers are x = 1 and x = −1. We have
f (−2) = −2/5, f (−1) = −1/2, f (1) = 1/2, f (2) = 2/5.
Since 1/2 > 2/5, the minimal and maximal values are −1/2 and 1/2 respectively.
4. Find the equation of the tangent line to the graph of f (x) = ln x at
x = 5.
Answer: y = 51 x + (ln 5 − 1).
5. For a given function:
• Find the domain
• Determine the equations of vertical and horizontal asymptotes
• Find the derivative and determine the intervals where the function is
increasing/decreasing
• Find the second derivative and determine the intervals where the function is concave up/down, find inflection points
• Draw the graph using all the information above
a) f (x) = 2x3 − 3x2 + 1
Solution: The function is defined everywhere, and there are no asymptotes. We have f 0 (x) = 6x2 − 6x = 6x(x − 1), so the function is increasing for
x > 1 and x < 0, and decreasing for 0 < x < 1. Furthermore, f 00 (x) = 12x−6,
so the function is concave up for x > 1/2 and concave down for x < 1/2, and
it has an inflection point at x = 1/2.
3
0
−10
−20
−30
−2
−1
0
1
2
b) f (x) = xe−x
Solution: The function is defined everywhere. To find the asymptotes,
remark that at x → −∞ the function e−x goes to infinity, so limx→−∞ xe−x =
−∞. To find the limit at x → +∞, let us use the L’Hôpital’s rule:
x
1
=
lim
= 0.
x→+∞ ex
x→+∞ ex
lim xe−x = lim
x→+∞
Therefore the graph has a horizontal asymptote y = 0 at x → +∞.
Now f 0 (x) = e−x − xe−x = (1 − x)e−x , so the function increases for
x < 1 and decreases for x > 1. Furthermore, f 00 (x) = −e−x − (1 − x)e−x =
(x − 2)e−x , so the function has an inflection point at x = 2.
0
−1
−2
−3
−1
0
1
2
3
4
c) f (x) = ln(x2 + 1)
4
Solution: Since x2 + 1 ≥ 1, the function is defined and nonnegative
everywhere. As x approaches ±∞, x2 + 1 → +∞, so ln(x2 + 1) → +∞, and
there are no horizontal asymptotes.
Now f 0 (x) = x22x+1 , so the function decreases for x < 0 and increases for
x > 0. Furthermore,
2(x2 + 1) − 2x(2x)
2(1 − x2 )
f (x) =
= 2
,
(x2 + 1)2
(x + 1)2
00
and the function has inflection points at x = ±1.
2.5
2
1.5
1
0.5
0
−3 −2 −1
d) f (x) =
0
1
2
3
x−1
x+1
Solution: The function is defined for x 6= −1, and has a vertical asymptote at x = −1. To find horizontal asymptotes, we have
x−1
1 − 1/x
= lim
= 1.
x→∞ x + 1
x→∞ 1 + 1/x
lim
2
Now f 0 (x) = x+1−(x−1)
= (x+1)
2 , so the function increases on every inter(x+1)2
val where it is defined. Furtermore, f 00 (x) = −4(x + 1)3 , so the function is
concave down for x > −1 and concave up for x < −1.
5
10
5
0
−5
−10
−3 −2 −1
0
1
2
3
6. Find
R 2x the integral:
a) e dx
Solution: Let u = 2x, then du = 2dx, dx = 1/2du, and
Z
Z
1
1
1
2x
e dx =
eu du = eu + C = e2x + C.
2
2
2
R
b) (x − 3)2 dx
Solution: Let u = x − 3, then du = dx and
Z
Z
u3
(x − 3)3
2
(x − 3) dx = u2 du =
+C =
+ C.
3
3
R
√
c) (x − 3)2 xdx
Solution: Let us expand (x − 3)2 :
Z
Z
Z
√
√
2
2
(x − 3) xdx = (x − 6x + 9) xdx = (x5/2 − 6x3/2 + 9x1/2 )dx =
2 7/2 12 5/2 18 3/2
x − x + x + C.
7
5
3
d)
R
x sin(x2 + 3)dx
Solution: Let u = x2 + 3, the du = 2xdx, and
Z
Z
1
1
1
2
x sin(x + 3)dx =
sin(u)du = − cos(u) + C = − cos(x2 + 3) + C.
2
2
2
6
e)
1
dx
x ln x
R
Solution: Let u = ln x, then du = x1 dx and
Z
Z
1
1
dx =
du = ln(u) + C = ln(ln x) + C.
x ln x
u
f)
2+x2
dx
1+x2
R
Solution: We have
Z
Z
Z
2 + x2
1 + x2 + 1
1
dx
=
dx
=
(1
+
)dx = x + arctan x + C.
1 + x2
1 + x2
1 + x2
7. Find
R 3 1the definite integral:
a) 0 1+x
2 dx
Solution: We have
Z 3
1
dx = arctan(x)|30 = arctan(3) − arctan(0) = arctan(3).
2
0 1+x
b)
R3
x
dx
0 1+x2
Solution: Let u = 1 + x2 , then du = 2xdx, so
Z 3
Z
x
1 51
1
ln(5)
dx =
du = (ln(5) − ln(1)) =
.
2
2 1 u
2
2
0 1+x
c)
R π2 /4
0
√
cos( x)
√
dx
x
Solution: Let u =
π 2 /4
Z
0
d)
R5
1
√
x, then du =
1
√
dx,
2 x
so
√1 dx
x
= 2du and
√
Z π/2
cos( x)
√
dx =
2 cos(u)du = 2(sin(π/2) − sin(0)) = 2.
x
0
eln x dx
Solution:
Z
5
ln x
e
1
Z
dx =
5
xdx =
1
7
x2 5 25 − 1
| =
= 12.
2 1
2
e)
R5
1
(ln x)5
dx
x
Solution: Let u = ln x, then du = x1 dx and
Z
1
5
(ln x)5
dx =
x
Z
ln 5
u5 du =
0
u6 ln 5 (ln 5)6
| =
.
6 0
6
8. Find the area
a) Between the parabola y = x2 and the line y = 3x − 2
Solution: Let us find the intersection points:
x2 = 3x − 2 ⇔ x2 − 3x + 2 = 0 ⇔ (x − 1)(x − 2) = 0,
so the intersection points are x = 1 and x = 2. Now
Z 2
3x2
x3
(3x−2−x2 )dx = (
A=
−2x− )|21 = 6−4−8/3−3/2+2+1/3 = 1/6.
2
3
1
b) Between the lines x = 0, y = x and y = 3 − 2x.
Solution: Let us find the intersection point between the second and the
third line:
x = 3 − 2x ⇔ 3x = 3 ⇔ x = 1.
Therefore
Z 1
Z 1
A=
(3 − 2x − x)dx =
(3 − 3x) = (3x − 3x2 /2)|10 = 3 − 3/2 = 3/2.
0
0
c) Between the hyperbola y = 1/x and the line y =
5
2
−x
Solution: Let us find the intersection points:
1
5
5
5
= − x ⇔ 1 = x − x2 ⇔ x2 − x + 1,
x
2
2
2
p
5/2 ± 25/4 − 4
5/2 ± 3/2
x1,2 =
=
,
2
2
so x1 = 1/2, x2 = 2.
8
Now
Z
A=
2
5
1
5
( − x − )dx = ( x − x2 /2 − ln(x))|21/2 =
x
2
1/2 2
5 − 2 − ln(2) − 5/4 + 1/8 + ln(1/2) = 15/8 − 2 ln(2).
d) Between the parabolas y = x2 − 3 and y = 5 − 3x2 .
Solution: Let us find the intersection points:
x2 − 3 = 5 − 3x2 ⇔ 4x2 = 8 ⇔ x2 = 2,
√
so the parabolas intersect at x = ± 2. We have
√
Z
A=
√
2
√
− 2
(5 − 3x2 − x2 + 3)dx =
Z
2
√
− 2
√
2
(8 − 4x2 ) = (8x − 4x3 /3)|−√
=
2
√
√
√
√
√
= 8 2 − 8 2/3 − 8(− 2) + 8(− 2)/3 = 32 2/3.
9. Find the volume of the body obtained by the rotation around x-axis
of the following region bounded by the lines x = 1, x = 4 and the graph of:
√
a) y = x
√
Solution: The sections
of
this
body
are
circles
with
radius
x, the area
√
of such a circle equals π( x)2 . We have:
Z 4
Z 4
√ 2
16 − 1
15π
V =
π( x) dx = π
xdx = πx2 /2|41 = π
=
.
2
2
1
1
b) y = x2
Solution: Similarly,
Z 4
Z
2 2
V =
π(x ) dx = π
1
4
x4 dx = πx5 /5|41 = π
1
c) y = x + 1
9
45 − 1
1023π
=
.
5
5
Solution: Similarly,
Z 4
Z 4
53 − 23
117π
2
(x + 1)2 dx = π(x + 1)3 /3|41 = π
π(x + 1) dx = π
=
.
V =
3
3
1
1
d) y = ex + 1
Solution: Similarly,
Z 4
Z 4
x
2
V =
π(e + 1) dx = π
(e2x + 2ex + 1)dx = π(1/2e2x + 2ex + x)|41 =
1
1
= π(e8 /2 + 2e4 + 4 − 1/2e2 − 2e − 1).
10. Consider the function
(
x + 1,
if x < −1
f (x) =
2
x + ax + b, if x ≥ −1.
a) For which values of the parameters it is continuous?
Solution: The function is continuous, if its limits from the left and from
the right at x = −1 coincide:
(−1) + 1 = (−1)2 + a(−1) + b ⇔ 0 = 1 − a + b ⇔ b = a − 1.
b) For which values of the parameters it has a derivative at every point?
Solution: The function clearly has a derivative for all x 6= −1, and it
has derivative at x = −1 if it is continuous at this point (see (a)), and the
derivatives from the left and from the right coincide: 1 = 2(−1) + a, so a = 3
and b = 2.
R3
c) For all a and b, find the integral −3 f (x)dx.
Solution:
Z
3
Z
−1
f (x)dx =
−3
−3
Z 3
3
2
3
(x+1)dx+
(x2 +ax+b)dx = (x2 /2+x)|−1
−3 +(x /3+ax /2+bx)|−1 =
−1
1/2 − 1 − 9/2 + 3 + 9 + 9/2a + 3b + 1/3 − 1/2a + b = 22/3 + 4a + 4b.
10