Solution: APPM 1350
Review #4
Summer 2014
1. Evaluate the following integrals:
√
Z
Z
Z
1
sin( x)
√
dx
(b)
(a)
dt
(c)
sec3 (x) tan(x)dx
(5t + 4)2.7
x
Z 1
Z
Z 1
Z 2
p
2
3 5
2
(x + 3|x|)dx
(g) 10
(d)
x (1 + 2x ) dx (e)
(x + 3) 4 − x dx (f)
−2
0
Solution:
(a) Let u =
√
−1
0
3
(1 + x5 ) /2 x9 dx
−1
√
x then du = dx/2 x and,
√
Z
Z
√
sin( x)
√
dx = 2 sin(u) du = −2 cos(u) + C = −2 cos( x) + C
x
(b) Let u = 5t + 4 then du = 5dt and so
Z
Z
1
1
dt =
(5t + 4)2.7
5
Z
u
sec3 (x) tan(x)dx =
(c)
−2.7
Z
1
du =
5
u−1.7
−1.7
(5t + 4) −1.7
−1
+C =−
+C
+C =
8.5
8.5(5t + 4)1.7
Z
sec2 (x) sec(x) tan(x)dx |{z}
=
u2 du =
u=sec(x)
Z
1
(d)
0
x2 (1 + 2x3 )5 dx |{z}
=
u3
sec3 (x)
+C =
+C
3
3
1 3 5
u6 3
u du =
= 728/36 = 182/9
6 1
36 1
Z
u=1+2x3
p
√
4 − x2 dx, and note that f (x) = x 4 − x2
−2
−2
−2
Z 2p
R2 √
2
4 − x2 dx geometrically as the
is an odd function so −2 x 4 − x dx = 0 and if we interpret
−2
Z 2p
area of a semicircle of radius 2 then
4 − x2 dx = 2π and thus,
Z
(e) Note that
2
2
Z
p
2
x 4 − x dx+
2
Z
p
2
x 4 − x dx + 3
2
Z
p
2
(x+3) 4 − x dx =
3
−2
Z
2
Z
p
2
(x + 3) 4 − x dx =
−2
2
−2
p
4 − x2 dx = 0 + 3(2π) = 6π
−2
(f)
Z
1
Z
0
(x + 3|x|)dx = −
−1
(g) If u = 1 +
Z
x5 ,
2x dx +
−1
then du =
5x4
0
10
Z
Z
5 3/2 9
x5
dx (and note
0
1
2
4x dx = −x + 2x = 3
2
−1
1
= 2
4
Z
f (x) dx = 10, find
0
5
3
(u /2 − u /2 ) du
0
2. If f is continuous and
1
Z
3/2
0
Z
0
= u − 1), so
u (u − 1) du = 2
(1 + x ) x dx = 2
−1
0
1
2 2
8
2 7/2 2 5/2 1
u − u
=2
−
=−
7
5
7 5
35
0
2
f (2x) dx
0
Solution: Let u = 2x, then du = 2dx and so
Z
Z 2
1 4
f (2x) dx =
f (u) du = 5.
2 0
0
√
3. Find the average value of f (t) = t 1 + t2 over the interval [0, 5].
Solution:
fave
1
=
5−0
5
Z
0
4. Find the limit:
√
(a) lim x x (b) lim
x→∞
x→1+
−6
x−5
Solution:
(e) lim
Z
p
1 1 26 √
1
2
t 1 + t dt |{z}
=
u du =
·
5 2 1
10
u=1+t2
1
1
−
ln(x) x − 1
(f) lim
x3
− 2x + 3
5x2 − 2x3
x→∞
x→5−
(c) lim x2 ex
x→−∞
!
2 3/2 26
1
u = (263/2 − 1)
3
15
1
2−x
x→1 (x − 1)2
(d) lim
x−4
x→4 |4 − x|
(g) lim
(a) Note that lim x1/x = “∞0 ”, and if we let y = lim x1/x then, by continuity,
x→∞
x→∞
1/x
1
ln(x) L0 H
= lim
= lim
=0
x→∞ 1
x→∞ x
x→∞ x
ln(y) = lim
√
x
So, ln(y) = 0 and so y = lim
x→∞
x = e0 = 1.
1
1
−
= “∞ − ∞” and so combining the two fractions and using
ln(x)
x−1
L’Hospital’s Rule twice yields,
(b) Note that lim
x→1+
lim
x→1+
1
1
x − 1 − ln(x) L0 H
1
1 − 1/x
1/x2
L0 H
−
= lim
=
= lim
= lim
2
+
+
+
ln(x) x − 1 x→1 (x − 1) ln(x)
2
x→1 ln(x) + 1 − 1/x
x→1 1/x + 1/x
(c) Note that lim x2 ex = “∞ · 0”, so, in the expression, if we divide one term by the reciprocal of
x→−∞
the other term and apply L’Hospitals Rule twice, this yields,
x2 L0 H
2x L0 H
2
= lim
= lim −x = 0
−x
−x
x→−∞ e
x→−∞ −e
x→−∞ e
lim x2 ex = lim
x→−∞
2−x
= +∞.
x→1 (x − 1)2
−6
(e) Here lim
= +∞.
x→5− x − 5
(d) Here we have lim
x3 − 2x + 3 DOP 1
(f) By dominance of powers, we have lim
= −
x→∞ 5x2 − 2x3
2
x−4
x−4
4 − x, if x ≤ 4
(g) Recall that |4 − x| =
thus, lim
= 1 and lim
= −1 and
+
−
−(4 − x), if x > 4
x→4 −(4 − x)
x→4 4 − x
x−4
so lim
= does not exist.
x→4 |4 − x|
5. Find y 0 :
(a) y = ln(sinh(x))
(b) y =
√
(e) y = 5−1/x
(f) y = xsin(x)
Z
2
2
(i) y = ln(x + y ) (j) y =
(c) y = (1 + x2 ) arctan(x) (d) y = x ln(tan−1 (x))
p
(g) y = log10 (x2 − 4) (h) y = 1 + 2e3x
xecosh(3x)
0
1/x2
Solution:
cosh(x)
(a) y 0 =
sinh(x)
(b) y 0 =
sin3 (t) dt
(k) y =
ln(x)
ex
ecosh(3x)
+ 3x1/2 sinh(3x)ecosh(3x)
2x1/2
(c) y 0 = 2x arctan(x) + 1
(d) y 0 = ln(tan−1 (x)) + x ·
(e) y 0 = 5−1/x · ln(5) ·
1
1
x
·
= ln(tan−1 (x)) +
−1
2
2
1 + x tan (x)
(1 + x ) tan−1 (x)
1
5−1/x ln(5)
=
x2
x2
y0
sin x
= cos(x) ln(x) +
y
x
sin x
sin x
= xsin(x) cos(x) ln(x) +
y 0 = y cos(x) ln(x) +
x
x
(f) Using logarithmic differentiation we have ln(y) = sin(x) ln(x), and so,
and thus,
ln(x2 − 4)
2x
, so, y 0 =
ln(10)
ln(10)(x2 − 4)
(g) Note that y = log10 (x2 − 4) =
(h) Note y = (1 + 2e3x )1/2 , so, y 0 = 3e3x (1 + 2e3x )−1/2
2x + 2yy 0
and so y 0 (x2 + y 2 ) = 2x + 2yy 0 , and thus
(i) Using implicit differentation, we have, y 0 = 2
x + y2
y0 =
Z
0
Z
3
sin (t) dt = −
(j) Note y =
1/x2
0
1/x2
x2
2x
+ y 2 − 2y
2
sin (t) dt, and by the FTOC we have, y = 3 sin3
x
0
3
1 x
e − ln(x)ex
1 − x ln(x)
x
(k) Here, y =
=
e2x
xex
√
6. Find the exact value of: (a) arcsin(−1/ 2)
1
x2
0
(b) cos(sin−1 (x))
Solution:
√
1
2
π
π
(a) If we let y = arcsin(−1/ 2), then sin(y) = − √ = −
, where − ≤ y ≤ , so y = −π/4,
2
2
2
2
√
that is arcsin(−1/ 2) = −π/4.
π
π
(b) If we let y = sin−1 (x), then sin(y) = x, where − ≤ y ≤ . Now note that cos2 (y) + sin2 (y) = 1
2
2
implies
q
p
cos(y) = ± 1 − sin2 (y) = ± 1 − x2
√
π
π
Now note that − ≤ y ≤ implies cos(y) ≥ 0 and so cos(sin−1 (x)) = cos(y) = 1 − x2 .
2
2
√
7. A bacteria culture initially contains 10 cells and grows at a rate proportional to its size. After an
hour the population has increased to 106. (a) Find an expression for the number of bacteria after t
hours. (b) When will the population reach 10,000 cells?
Solution:
(a) Let y(t) be the population of the bacteria at t hours, then y(t) = y0 ekt and y(0) = 10 implies
y0 = 10. Now y(1) = 106 implies 106 = 10ek , and so k = ln(10.6) and thus y(t) = 10eln(10.6)t
(b) We wish to find t such that y(t) = 10, 000, or 10eln(10.6)t = 10, 000, which implies eln(10.6)t = 1000
ln(1000)
and so ln(10.6)t = ln(1000) or t =
.
ln(10.6)
√
1− x
√
8. (a) Find f −1 given f (x) =
(b) Find (f −1 )0 (1) given f (x) = x3 + x + 1.
1+ x
Solution:
1−x 2
−1
(a) f (x) =
1+x
(b) Note that f (0) = 1 and so f −1 (1) = 0 and recall (f −1 )0 (a) =
1
f 0 (f −1 (a))
, and here f 0 (x) = 3x2 +1
and so,
(f −1 )0 (1) =
1
=
f 0 (f −1 (1))
1
f 0 (0)
=
1
=1
1
9. Evaluate the integral:
Z
Z
Z
Z
dx
ex
dx
√
(c)
sec(x) dx (d)
(a)
dx (b)
2x
e +1
sec(x) + tan(x)
x+ x
Z
Z 1
Z
dx
x
√
dx (g)
(e)
e|x| dx (f)
2
x cos(ln(x))
x +1
−1
Solution:
Z
Z
ex
1
x
(a) If we let u = e , then,
dx =
du = tan−1 (u) + C = tan−1 (ex ) + C
2x
2
e +1
u +1
Z
Z
√
dx
dx
dx
√ =
√ √
(b) Note
and if we let u = x + 1 then 2du = √ and so
x+ x
x( x + 1)
x
Z
dx
√ =
x+ x
Z
√
u= x+1 Z
dx
z}|{
√ √
=
x( x + 1)
√
2du
= 2 ln |u| + C = 2 ln | x + 1| + C
u
(c) Note, if we multiply the top and bottom by sec(x) + tan(x), then we have
Z
Z
Z
sec2 (x) + sec(x) tan(x)
sec(x) + tan(x)
dx =
dx
sec(x) dx = sec(x) ·
sec(x) + tan(x)
sec(x) + tan(x)
and, now, if we let u = sec(x) + tan(x), then du = (sec2 (x) + sec(x) tan(x)) dx, and so,
Z
Z
sec(x) dx =
sec2 (x) + sec(x) tan(x)
dx
sec(x) + tan(x)
u=sec(x)+tan(x) Z
z}|{
=
du
= ln |u|+C = ln | sec(x)+tan(x)|+C
u
(d) Note, if we multiply the top and bottom by cos(x), then we have
Z
Z
Z
1
1
cos(x)
cos(x)
dx =
dx =
dx
sec(x) + tan(x)
sec(x) + tan(x) cos(x)
1 + sin(x)
now if we let u = 1 + sin(x) then we get du = cos(x), and so,
Z
Z
1
dx =
sec(x) + tan(x)
cos(x)
dx
1 + sin(x)
u=1+sin(x) Z
z}|{
=
du
= ln |u| + C = ln |1 + sin(x)| + C
u
0
−x 1
x
(e) Note that,
Z
1
e
|x|
Z
0
dx =
−1
−x
e
Z
(f) If we let u = x2 + 1 then
x
e dx = −e
dx +
−1
1
0
Z
√
−1
1
x
dx =
2
x2 + 1
Z
+ e = (−1 + e) + (e − 1) = 2(e − 1)
0
p
du
1
√ = · 2u1/2 + C = x2 + 1 + C
2
u
(g) If we let u = ln(x) then
Z
dx
=
x cos(ln(x))
Z
du
=
cos(u)
from part (c)
Z
sec(u) du
z}|{
=
ln | sec(u)+tan(u)|+C = ln |sec(ln(x)) + tan(ln(x))|+C
Z
2
10. (a) Find an approximation to the integral
x2 dx using a Riemann sum with a regular partition
1
with left endpoints and n = 4.
(b) Given that
n
X
i=1
n
X
n(n + 1)
i=
,
2
i=1
n(n + 1)(2n + 1)
,
i =
6
2
and
n
X
i=1
n(n + 1)
i =
2
3
2
.
Use the limit
definition of the integral, with right endpoints and a regular partition, to find the exact
Z
2
x2 dx.
value of
1
Solution:
(a) Here, ∆x = (b − a)/n = 1/4 and xi−1 = a + (i − 1)∆x = 1 + (i − 1)/4 = (i + 3)/4, thus
Z
2
2
x dx ≈
1
4
X
x∗i =xi−1
f (x∗i )∆xi
z}|{
=
i=1
=
=
4
X
i+3 1
f
4
4
i=1
5
6
7
1
f (1) + f
+f
+f
4
4
4
4
1
25 36 49
1 126
63
1+
+
+
=
=
4
16 16 16
4 16
32
(b) Note that ∆x = (2 − 1)/n = 1/n and xi = 1 + i∆x = 1 + i/n and so,
Z 2
n
n X
X
i 21
2
x dx |{z}
= lim
f (xi )∆x = lim
1+
n→∞
n→∞
n
n
1
i=1
i=1
x∗i =xi
" n
#
n
n
1X
2 X
1 X 2
= lim
1+ 2
i+ 3
i
n→∞ n
n
n
i=1
i=1
i=1
1
2 n(n + 1)
1 n(n + 1)(2n + 1)
= lim
n+ 2
+ 3
n→∞ n
n
2
n
6
3
2
1
7
2n + 2n 2n + (lower order terms)
+
=1+1+ =
= lim 1 +
2
3
n→∞
2n
6n
3
3
11. If 1200 cm2 of material is available to make a box with a square base and an open top, find the
largest possible volume of the box.
Solution:
Let x be the length of the side of the base of the box and let h be the height of the box. Then we wish
to maximize the volume, V = x2 h, subject to the constraint on the surface area, x2 + 4xh = 1200.
2
The constraint implies
= x(1200 − x2 )/4 = 300x − x3 /4, and h ≥ 0
√ h = (1200 − x0 )/4x and so V (x)
2
implies 0 ≤ x ≤ 20 3. Note that V (x) = 300 − 3x /4 and V 0 (x) = 0 implies x = ±20 and so the
00
critical point is x = 20. Note that V 00 (x) =
√ −6x/4 and so V (20) < 0 which implies x = 20 is a
local max. Finally, note that V (0) = V (20 3) = 0 and V (20) = 4000 and so we have an absolute
maximum at x = 20 and so the largest possible volume of the box is 4000 cm3 .
12. Complete two iterations of Newton’s Method to approximate a root of y = 3x2 − 2 using an initial
guess of x1 = 1.
Solution:
Note f 0 (x) = 6x and after the first iteration, we have
x2 = x1 −
f (x1 )
3x21 − 2
3(1)2 − 2
1
5
=
x
−
=
1
−
=1− =
1
0
f (x1 )
6x1
6(1)
6
6
then after the second iteration we have
x3 = x2 −
f (x2 )
5 3(5/6)2 − 2
3x22 − 2
5 25/12 − 2
5
1
49
=
=
x
−
−
= −
= −
=
2
0
f (x2 )
6x2
6
6(5/6)
6
5
6 60
60
so after two iterations we have x3 = 49/60.
2
3x5/3
(a) State the domain of the function, does the function have any vertical or horizontal asymptotes?
13. Given that y = x − 3x1/3 , y 0 = 1 −
1
x2/3
, y 00 =
(b) Determine the intervals where y is increasing and decreasing.
(c) Determine the intervals where y is concave up and concave down.
(d) Find any local maximum or minimum values.
(e) Sketch the graph of the function.
Solution:
(a) The domain is −∞ < x < ∞. There are no vertical or horizontal asymptotes.
(b) Increasing on (−∞, −1) ∪ (1, ∞) and decreasing on (−1, 0) ∪ (0, 1) (or (−1, 1) is acceptable as
well). Note that there is a vertical tangent at x = 0.
Sign Chart for y 0
%
&
&
|
-1
%
|
0
|
1
(c) Concave down on (−∞, 0) and concave up on (0, ∞)
Sign Chart for y 00
_
^
|
0
(d) Local max at (−1, 2) and local min at (1, −2)
√
(e) The graph, (note the x-intercepts are (± 27, 0) and (0, 0),)
2.4
1.6
0.8
-6
-5
-4
-3
-2
-1
0
-0.8
-1.6
-2.4
1
2
3
4
5
6
14. Given that f (x) =
x2 − x
2
−4
, f 0 (x) =
, f 00 (x) =
2
2
x +x−2
(x + 2)
(x + 2)3
(a) State the domain of the function, does the function have any vertical or horizontal asymptotes?
(b) Determine the intervals where y is increasing and decreasing.
(c) Determine the intervals where y is concave up and concave down.
(d) Find any local maximum or minimum values.
(e) Sketch the graph of the function.
Solution:
(a) Domain: x 6= 1, −2, vertical asymptote is x = −2 and removable discontinuity at (1, 1/3),
horizontal asymptote is y = 1
(b) The function is an increasing function for all vaues of x.
(c) Concave up on (−∞, −2) and concave down on (−2, 1) ∪ (1, ∞)
Sign Chart for y 00
^
_
_
|
-2
|
1
(d) There are no local extrema.
(e) The graph,
3
2
1
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
-1
-2
-3
15. (a) State the Intermediate Value Theorem
(b) State Rolle’s Theorem.
(c) State the Mean Value Theorem.
(d) Find all numbers c that satisfy the conclusion of the Mean Value Theorem for f (x) = ln(x) on
the interval [e, e2 ]
Solution:
(a) Intermediate Value Theorem: If f (x) is continuous on [a, b] and if N is any number between f (a)
and f (b), where f (a) 6= f (b), then there exist a number c in (a, b) such that f (c) = N .
(b) Rolle’s Theorem: If f (x) is continuous on [a, b] and differentiable on (a, b) and if f (a) = f (b)
then there exist a number c in (a, b) such that f 0 (c) = 0.
(c) Mean Value Theorem: If f (x) is continuous on [a, b] and differentiable on (a, b) then there exists
f (b) − f (a)
a number c in (a, b) such that f 0 (c) =
.
b−a
(d) We wish to find c such that f 0 (c) =
f (e2 ) − f (e)
, that is
e2 − e
1
2−1
= 2
and so, c = e2 − e.
c
e −e
(Note that e > 0 implies e2 −e < e2 and e > 2 implies e2 > 2e and so e2 −e > e, that is, e < e2 − e < e2 .)
16. Find the absolute maximum and absolute minimum values of f (x) = cosh(x) over the interval [ln(1/2), ln(3)].
Solution:
Note that f 0 (x) = sinh(x) and f 0 (x) = 0 implies x = 0. Now note that f (0) = 1 and f (ln(1/2)) = 5/4
and f (ln(3)) = 5/3 and so the absolute maximum is (ln(3), 5/3) and the absolute minimum is (0, 1)
17. A balloon is rising at a constant rate of 5 ft/s. A boy is cycling along a straight road at a speed of
15 ft/s. When he passes under the balloon, it is 45 ft above him. How fast is the distance between
the boy and the balloon increasing 3 seconds later?
Solution:
Let z be the distance between the balloon and the boy, let y be the height of the balloon and let x
dy
be the distance travelled by the boy from the spot below the balloon. So we have
= 5 ft/s and
dt
dx
= 15 ft/s, and we wish to find dz/dt when t = 3. Note that when t = 3, we have x = 45ft and
dt
y = 60ft and x2 + y 2 = z 2 implies z = 75ft. Differentiating the equation x2 + y 2 = z 2 with respect
dx
dy
dz
dz
45
60
975
to t yields 2x
+ 2y
= 2z , and so
= (15) + (5) =
= 13ft/s.
dt
dt
dt
dt
75
75
75
18. Find the linearization of f (x) = ex at the point a = 1/2, and then use this linearization to approximate e. (Hint: ln(1.6) ≈ 1/2)
Solution:
Note that L(x) = f (a) + f 0 (a)(x − a) and here a = 1/2, f 0 (1/2) = e1/2 , now from the hint we know
e1/2 ≈ 1.6, so the linearization is L(x) = e1/2 + e1/2 (x − 1/2) and note that
e = f (1) ≈ L(1) = 1.6 + 1.6(1 − 1/2) = 1.6 + 1.6(1/2) = 2.4
so based on this linearization e ≈ 2.4.
19. If a particle has position function s(t) = t3 − 12t + 3, where t ≥ 0 is measured in seconds and s in
feet.
(a) Find the velocity and acceleration function.
(b) When is the particle moving forward and when is it moving backward?
(c) Find the total distance traveled in the first 3 seconds.
Solution:
(a) v(t) = 3t2 − 12 and a(t) = 6t
(b) The particle is moving backward for 0 ≤ t < 2 and is moving forward for t > 2.
(c) Note
Z
3
Z
|v(t)| dt =
0
2
2
Z
−(3t − 12) dt +
0
3
(3t2 − 12) dt = 16 + 7 = 23.
2
20. Use the limit definition of the derivative to find f 0 (x) given f (x) =
Solution:
√
1 + 2x.
f (x + h) − f (x)
f (x) = lim
h→0
h
0
p
√
1 + 2(x + h) − 1 + 2x
= lim
h→0
h
p
p
√
√
1 + 2(x + h) − 1 + 2x
1 + 2(x + h) + 1 + 2x
= lim
·p
√
h→0
h
1 + 2(x + h) + 1 + 2x
1 + 2(x + h) − (1 + 2x)
2
= lim p
= lim p
√
√
h→0
h→0
1 + 2(x + h) + 1 + 2x
h
1 + 2(x + h) + 1 + 2x
=
√
1
1 + 2x
21. Find a and b so that
f (x) =
ax3 , x ≤ 2
x2 + b, x > 2
is differentiable everywhere.
Solution:
Note that we need continuity for differentiability to hold, now for continuity at x = 2 we need
lim f (x) = f (2) that is we need 4 + b = 8a and for differentiability we need lim f 0 (x) = lim f 0 (x),
x→2+
x→2+
that is we need 4 = 12a, so we need a = 1/3 which implies b = −4/3.
22. Given f (x) =
√
1
and g(x) = x find g/f , f ◦ f and g ◦ f and state their domains.
2
x
Solution:
√
(∗) Note that (g/f )(x) = x2 x and the domain is x > 0.
(∗) Note that (f ◦ f )(x) = x4 and the domain is x 6= 0.
1
(∗) Note that (g ◦ f )(x) = and the domain is x 6= 0.
x
x→2−